\documentclass[a4paper,11pt]{article}
\usepackage[latin1]{inputenc}
\def\ratio#1#2{\frac{\overline{#1}}{\overline{#2}}}
\newtheorem{geothm}{Theorem}
\pagestyle{empty}
\addtolength{\textwidth}{2cm}

\begin{document}

Nome Utilizador: gpreda1


Nome do Teorema: 
 Harmonic Points



Teorema: \begin{geothm}
 Given four points $A$, $B$, $C$ and $D$,
 points of intersections are formed: $L = AB \cap CD$, 
 $K = AC \cap BC$, $F = KL \cap BD$ and $G = KL \cap AC$.
 Show that segment $KL$ is harmonically divided by
 the points $F$ and $G$, ie:
 $\frac{\overline{LK}}{\overline{KF}} = \frac{\overline{GL}}{\overline{GF}}$.
\end{geothm}

Descrição Breve: 


Observações: 
dim  130 100
point  A 52 20
point  B 72 20
point  C 66 40
point  D 56 36
cmark_b  A
cmark_b  B
cmark_r  C
cmark_lt  D
line  AB A B
line  CD C D
intersec  L AB CD
cmark_b  L
line  AD A D
line  BC B C
intersec  K AD BC
cmark_lt  K
line  BD B D
line  KL K L
intersec  F BD KL
cmark_lt  F
line  AC A C
intersec  G AC KL
cmark_lt  G
drawsegment  B L
drawsegment  C L
drawsegment  A K
drawsegment  B K
drawsegment  B F
drawsegment  A G
line  lkfg L K
drawdashline  lkfg



Código GCLC: 
point  A 52 20
point  B 72 20
point  C 66 40
point  D 56 36
line  AB A B
line  CD C D
intersec  L AB CD
line  AD A D
line  BC B C
intersec  K AD BC
line  BD B D
line  KL K L
intersec  F BD KL
line  AC A C
intersec  G AC KL
prove { harmonic L K F G }



Código GCLCprover: 


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Thales  Theorem



Nome do Teorema: \begin{geothm}
 Two lines, $AB$ and $CD$ intersect at a point $O$.
 If $AC \parallel DB$, show that:
 $\frac{\overline{OA}}{\overline{OB}} = 
 \frac{\overline{OC}}{\overline{OD}}$
\end{geothm}

Teorema: 


Descrição Breve: 
dim  80 80
point  O 20 30 
point  A 60 30      
point  C 40 50  
point  B 35 30
line  a A C
line  c O C 
parallel  b B a
intersec  D c b 
drawline  O A 
drawline  O C 
drawline  A C
drawline  B D
cmark_t  O
cmark_t  A
cmark_t  B
cmark_t  C   
cmark_t  D   



Observações: 
point  O 20 30 
point  A 60 30      
point  C 40 50  
online  B  O A
line  a A C
line  c O C 
parallel  b  B  a
intersec  D  c b 
prove { equal { sratio O A O B  } { sratio O C O D  } }



Código GCLC: 


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Parallelogram Theorem



Nome Utilizador: \begin{geothm}
 Given triangle $ABC$, let $p$ be line through $A$ parallel
 with $BC$, and let $q$ be line through $C$ parallel
 with $AB$. Let $p$ and $q$ intersects at $D$. Show that
 $\overline{AB} = \overline{DC}$.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  A 20 20
point  B 80 20
point  C 103.40 55.10
line  ab A B
line  bc B C
parallel  p A bc
parallel  q C ab
intersec  D p q
prove { equal { sratio A B D C } { 1 } }


Observações: 


Código GCLC: gpreda1


Código GCLCprover: 
 Example 35 


Referência BiBTeX: \begin{geothm}
 Let $ABC$ be an equilateral triangle inscribed in a circle
 with center $O$, and let $P$ be any point on the circle.
 Then the Simson line of $P$ bisects the radius $OP$.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  130 110
point  A 30 30
point  B 100 30
circle  k1 A B
med  mc A B
intersec2  C C1 k1 mc
med  mb A C
intersec  O mc mb
circle  k O A
point  P 105.25 53.81
line  cb C B
line  ac A C
foot  D P cb
foot  E P ac
line  de D E
line  po P O
intersec  I po de
cmark_b  I
drawline  de
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  O
cmark_r  P
cmark_rt  D
cmark_l  E
drawdashsegment  P O
drawcircle  k
drawsegment  A B
drawsegment  A C
drawsegment  C B



Descrição Breve: chou88


Observações: gpreda1


Código GCLC: 
 Example 3 


Código GCLCprover: \begin{geothm}
 Starting from five points $A$, $B$, $C$, $D$ and $E$
 with $A$, $B$, $C$ collinear, new lines and points
 of intersection are formed. $ED$, $IG$, $LK$ and $JH$
 are collinear.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  100 100
point  A 20 20
point  B 55 20
line  ab A B
point  C 50 20
point  D 40.20 47.20
point  E 53.30 37.50
line  ae A E
line  cd C D
line  ad A D
line  be B E
line  bd B D
line  ec E C
line  de D E
intersec  J ae cd
intersec  G ad be
intersec  K bd ec
intersec  I ae bd
intersec  L cd be
intersec  H ec ad
intersec  F de ab
line  jh J H
line  ig I G
intersec  O jh ig
cmark_rb  A
cmark_b  B
cmark_lb  C
cmark_t  D
cmark_rb  E
cmark_lb  J
cmark_lt  G
cmark_r  K
cmark_b  I
cmark_r  L
cmark_rb  H
cmark_b  F
cmark_r  O
line  oed O E
line  olk O L
drawline  oed
drawline  olk
drawdashline  ab
drawdashline  ae
drawdashline  cd
drawdashline  ad
drawdashline  ec
drawdashsegment  I G
drawdashsegment  J H



Teorema: 
point  A 20 20
point  B 55 20
line  ab A B
online  C  A B
point  D 40.20 47.20
point  E 53.30 37.50
line  ae A E
line  cd C D
line  ad A D
line  be B E
line  bd B D
line  ec E C
line  de D E
intersec  J ae cd
intersec  G ad be
intersec  K bd ec
intersec  I ae bd
intersec  L cd be
intersec  H ec ad
intersec  F de ab
line  jh J H
line  ig I G
intersec  O jh ig
prove { collinear O E D }
prove { collinear O L K }


Descrição Breve: gpreda1


Observações: 
 Brahmagupta s Theorem



Código GCLC: \begin{geothm}
 In a cyclic quadrilateral having perpendicular 
 diagonals, the perpendicular to a side from 
 the point of intersection of the diagonals 
 always bisects the opposite side.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  110 110
point  B 20 40
point  T 43 40
point  O 55 60
circle  k O B
line  bd B T
perp  ac T bd
perp  ob O bd
intersec  X ob bd
towards  D B X 2
intersec2  A C1 k ac
perp  oa O ac
intersec  Y oa ac
towards  C A Y 2
line  ad A D
line  bc B C
perp  tp T bc
intersec  P tp ad
intersec  Q tp bc
cmark_rt  P
cmark_t  Q
cmark_l  B
cmark_r  D
cmark_t  A
cmark_b  C
cmark_b  O
cmark_rb  T
cmark_b  X
cmark_l  Y
drawdashsegment  P Q
drawsegment  A B
drawsegment  C B
drawsegment  C D
drawsegment  A D
drawsegment  A C
drawsegment  D B
drawcircle  k



Nome do Teorema: 
point  B 20 40
point  T 43 40
point  O 55 60
circle  k O B
line  bd B T
perp  ac T bd
perp  ob O bd
intersec  X ob bd
towards  D B X 2
intersec2  A C1 k ac
perp  oa O ac
intersec  Y oa ac
towards  C A Y 2
line  ad A D
line  bc B C
perp  tp T bc
intersec  P tp ad
intersec  Q tp bc
prove { equal { sratio A P P D } { 1 } }


Teorema: 


Descrição Breve: 
 Dual Altitude Theorem



Observações: \begin{geothm}
 Triangle $ABC$ and point $O$ are given.
 Let $p$, $q$ and $r$ be lines through the point
 $O$ which are perpendicular with lines $OA$, $OB$ and
 $OC$ respectively. Let $D$, $E$, and $F$ be
 intersections of lines $p$, $q$ and $r$ with lines
 $BC$, $AC$ and $BC$ respectively. Show that
 $D$, $E$ and $F$ are collinear.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  100 80
point  A 20 20
point  B 80 20
point  C 40 60
point  O 75 45
line  bc B C
line  ca C A
line  ab A B
line  oa O A
perp  od O oa
intersec  D od bc
line  ob O B
perp  oe O ob
intersec  E oe ca
line  oc O C
perp  of O oc
intersec  F of ab
cmark_b  A
cmark_b  B
cmark_t  C
cmark_r  O
cmark_b  D
cmark_t  E
cmark_b  F
drawsegment  A B
drawsegment  D C
drawsegment  C A
drawsegment  O A
drawsegment  O B
drawsegment  O C
drawdashsegment  O E
drawdashsegment  O F
drawdashsegment  O D
line  def D E
drawdashline  def



Nome Utilizador: 
point  A 20 20
point  B 80 20
point  C 40 60
point  O 75 45
line  bc B C
line  ca C A
line  ab A B
line  oa O A
perp  od O oa
intersec  D od bc
line  ob O B
perp  oe O ob
intersec  E oe ca
line  oc O C
perp  of O oc
intersec  F of ab
line  def D E
prove { collinear D E F }


Nome do Teorema: 


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Given points $O$, $A$, $B$, $C$ and $O_1$
 new points are constructed as:
 $A_1 = O_1B \cap OC$, $B_1 = OA \cap O_1C$,
 $C_1 = OB \cap O_1A$, $I = BB_1 \cap AA_1$.
 Show that points $C$, $C_1$ and $I$ are collinear.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  130 130
point  O 25.80 48.30
point  A 95.20 55.00
point  B 120.10 91.60
point  C 44.50 112.70
point  O_1 62.40 56.80
line  o1b O_1 B
line  oc O C
line  oa O A
line  o1c O_1 C
line  ob O B
line  o1a O_1 A
intersec  A_1 o1b oc
intersec  B_1 oa o1c
line  bb1 B B_1
line  aa1 A A_1
intersec  C_1 ob o1a
intersec  I bb1 aa1
line  cc1i C I
drawdashline  cc1i
cmark_b  A_1
cmark_rb  B_1
cmark_lt  C_1
cmark_rb  I
cmark_l  O
cmark_b  A
cmark_b  B
cmark_l  C
cmark_lt  O_1
drawsegment  A_1 C
drawsegment  O B
drawsegment  O A
drawsegment  A C_1
drawsegment  A_1 B
drawsegment  C B_1
drawdashsegment  A_1 A
drawdashsegment  I B



Referência BiBTeX: 
point  O 25.80 48.30
point  A 95.20 55.00
point  B 120.10 91.60
point  C 44.50 112.70
point  O_1 62.40 56.80
line  o1b O_1 B
line  oc O C
line  oa O A
line  o1c O_1 C
line  ob O B
line  o1a O_1 A
intersec  A_1 o1b oc
intersec  B_1 oa o1c
line  bb1 B B_1
line  aa1 A A_1
intersec  C_1 ob o1a
intersec  I bb1 aa1
line  cc1i C I
prove { collinear C C_1 I }


Nome Utilizador: 


Nome do Teorema: gpreda1


Teorema: 
 Example 46 


Descrição Breve: 


Observações: 


Código GCLC: 
dim  130 100
point  A 20 20
point  E 80 20
point  B 95 70
point  C 35 20
point  F 65 70
line  ae A E
parallel  bd B ae
line  be B E
parallel  ad A be
intersec  D bd ad
line  cd C D
line  af A F
intersec  M cd af
line  ef E F
line  bc B C
intersec  N ef bc
line  mn M N
intersec  P mn ad
intersec  Q mn be
line  pq P Q
cmark_b  A
cmark_b  E
cmark_t  B
cmark_t  D
cmark_b  C
cmark_t  F
cmark_rb  M
cmark_rb  N
cmark_lt  P
cmark_rb  Q
drawsegment  A E
drawsegment  E B
drawsegment  B D
drawsegment  D A
drawdashsegment  D C
drawdashsegment  A F
drawdashsegment  E F
drawdashsegment  C B
drawline  pq



Código GCLCprover: 
point  A 20 20
point  E 80 20
point  B 95 70
point  C 35 20
point  F 65 70
line  ae A E
parallel  bd B ae
line  be B E
parallel  ad A be
intersec  D bd ad
line  cd C D
line  af A F
intersec  M cd af
line  ef E F
line  bc B C
intersec  N ef bc
line  mn M N
intersec  P mn ad
intersec  Q mn be
line  pq P Q
prove { equal { sratio A P Q B } { 1 } }


Referência BiBTeX: chou88


Nome Utilizador: gpreda1


Nome do Teorema: 
 Example 49 


Teorema: \begin{geothm}
 Prove that the line joining the point of intersection
 of the extensions of the nonparallel sides of a trapezoid
 to the point of intersection of its diagonals bisects the base
 of the trapezoid.
\end{geothm}

Descrição Breve: 


Observações: 
dim  130 80
point  A 20 20
point  B 110 20
point  C 90 50
point  D 55 50
line  bc B C
line  ad A D
line  cd C D
line  ab A B
line  ac A C
line  bd B D
intersec  F ad bc
intersec  E ac bd
line  ef E F
intersec  G ab ef
cmark_b  A
cmark_b  B
cmark_rt  C
cmark_lt  D
cmark_rb  E
cmark_rt  F
cmark_b  G
drawsegment  A F
drawsegment  B F
drawsegment  G F
drawsegment  A B
drawsegment  C D
drawsegment  A C
drawsegment  B D



Código GCLC: 
point  A 20 20
point  B 110 20
point  C 90 50
point  D 55 50
line  bc B C
line  ad A D
line  cd C D
line  ab A B
line  ac A C
line  bd B D
intersec  F ad bc
intersec  E ac bd
line  ef E F
intersec  G ab ef
prove { equal { sratio A G G B } { 1 } }



Código GCLCprover: chou88


Referência BiBTeX: gpreda1


Nome Utilizador: 
theorems\thm_0070_Altitude.gcl


Nome do Teorema: \begin{geothm}
 Show that triangle altitudes are concurrent.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  100 85
point  A 20 20
point  B 80 20
point  C 60 70
line  bc B C
line  ab A B
line  ac A C
foot  A_1 A bc
foot  B_1 B ac
foot  C_1 C ab
line  ha A A_1
line  hb B B_1
line  hc C C_1
intersec  H hc hb
cmark_b  A
cmark_b  B
cmark_t  C
cmark_rt  A_1
cmark_l  B_1
cmark_b  C_1
cmark_lb  H
drawsegment  A B
drawsegment  A C
drawsegment  B C
drawdashsegment  B B_1
drawdashsegment  A A_1
drawdashsegment  C C_1



Observações: 
point  A 20 20
point  B 80 20
point  C 60 70
line  bc B C
line  ab A B
line  ac A C
foot  A_1 A bc
foot  B_1 B ac
foot  C_1 C ab
line  ha A A_1
line  hb B B_1
line  hc C C_1
intersec  H hc hb
prove { collinear  A H A_1 }



Código GCLC: 


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Example 56 


Nome Utilizador: \begin{geothm}
 Let $l$ be a line passing through the vertex
 of $M$ of a parallelogram $MNPQ$ and intersecting
 the lines $NP$, $PQ$, $NQ$ in points $R$, $S$, $T$.
 Show that $1/MR + 1/MS = 1/MT$.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  M 20 20
point  N 70 20
point  P 85 45
line  mn M N
line  np N P
parallel  pq P mn
parallel  mq M np
intersec  Q pq mq
point  S 120 45
line  ms M S
line  nq N Q
intersec  T ms nq
intersec  R np ms
prove { equal { sum { sratio M T M R } { sratio M T M S } } { 1 } } }


Observações: chou88


Código GCLC: gpreda1


Código GCLCprover: 
 Example 58 


Referência BiBTeX: \begin{geothm}
 Let $A_1$, $B_1$, $C_1$ be points on the sides
 $BC$, $CA$, $AB$ of a triangle $ABC$ such that
 $BA_1/BC = CB_1/CA_1 = AC_1/AB = 1/3$. Show
 that the area of the triangle determined by lines
 $AA_1$, $BB_1$ and $CC_1$ is one seventh of the area
 of triangle $ABC$.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  130 100
point  A 20 20
point  B 90 20
towards  C_1 A B 0.333333333
point  C 70 80
towards  A_1 B C 0.333333333
towards  B_1 C A 0.333333333
line  aa1 A A_1
line  bb1 B B_1
line  cc1 C C_1
intersec  A_2 bb1 cc1
intersec  B_2 aa1 cc1
intersec  C_2 aa1 bb1
cmark_b  A
cmark_b  B
cmark_b  C_1
cmark_r  C
cmark_r  A_1
cmark_lt  B_1
cmark_b  C_2
cmark_rb  B_2
cmark_r  A_2
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C C_1
drawsegment  A A_1
drawsegment  B B_1



Descrição Breve: chou88


Observações: gpreda1


Código GCLC: 
 Triangle Area



Código GCLCprover: \begin{geothm}
 Show that median $CM$ divides triangle
 $ABC$ into two triangles of the same
 area.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  100 70
point  A 20 20
point  B 80 20
point  C 65 45
midpoint  M A B
cmark_b  A
cmark_b  B
cmark_t  C
cmark_b  M
drawsegment  A B
drawsegment  B C
drawsegment  A C
drawsegment  C M



Teorema: 
point  A 20 20
point  B 80 20
point  C 65 45
midpoint  M A B
prove { equal { signed_area3 A M C } { signed_area3 M B C } }


Descrição Breve: gpreda1


Observações: 
 Example 55 


Código GCLC: \begin{geothm}
 Let $M$, $N$, $P$ be points on the sides $AB$, $BC$ and $AC$
 of a triangle $ABC$. Show that if $M_1$, $N_1$ and $P_1$
 are points on sides $AC$, $BA$ and $BC$ of a triangle
 $ABC$ such that $MM_1 \parallel BC$, $NN_1 \parallel CA$
 and $PP_1 \parallel AB$, then triangles $MNP$ and
 $M_1N_1P_1$ have equal areas.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  130 110
point  A 20 20
point  B 110 20
point  C 80 90
point  M 88 20
line  ab A B
line  ac A C
line  bc B C
point  N 97.62 48.88
point  P 36.68 39.46
parallel  pp1 P ab
parallel  nn1 N ac
parallel  mm1 M bc
intersec  P_1 pp1 bc
intersec  M_1 mm1 ac
intersec  N_1 nn1 ab
cmark_b  A
cmark_b  B
cmark_t  C
cmark_rt  N
cmark_b  M
cmark_lt  P
cmark_rt  P_1
cmark_b  N_1
cmark_lt  M_1
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  P P_1
drawsegment  M M_1
drawsegment  N N_1
drawdashsegment  M N
drawdashsegment  M P
drawdashsegment  P N
drawdashsegment  M_1 N_1
drawdashsegment  M_1 P_1
drawdashsegment  P_1 N_1



Nome do Teorema: 
point  A 20 20
point  B 110 20
point  C 80 90
point  M 88 20
line  ab A B
line  ac A C
line  bc B C
online  N  B C
online  P  A C
parallel  pp1 P ab
parallel  nn1 N ac
parallel  mm1 M bc
intersec  P_1 pp1 bc
intersec  M_1 mm1 ac
intersec  N_1 nn1 ab
prove { equal { signed_area3 M N P } { signed_area3 M_1 N_1 P_1 } }


Teorema: chou88


Descrição Breve: 
 Example 57 


Observações: \begin{geothm}
 Let $A_1$, $B_1$, $C_1$, $D_1$ be points on the sides $CD$,
 $DA$, $AB$, $BC$ of a parallelogram $ABCD$ such that $CA_1/CD = DB_1/DA = AC_1/AB = BD_1/BC = 1/3$.
 Show that the area of the quadrilateral formed by the lines
 $AA_1$, $BB_1$, $CC_1$, $DD_1$ is one thirteenth of the area of
 parallelogram $ABCD$.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  130 100
point  A 20 20
point  B 90 20
point  C 110 80
line  ab A B
line  bc B C
parallel  ad A bc
parallel  cd C ab
intersec  D ad cd
towards  C_1 A B 0.333333333
towards  D_1 B C 0.333333333
towards  A_1 C D 0.333333333
towards  B_1 D A 0.333333333
line  aa1 A A_1
line  bb1 B B_1
line  cc1 C C_1
line  dd1 D D_1
intersec  A_2 aa1 bb1
intersec  B_2 bb1 cc1
intersec  C_2 dd1 cc1
intersec  D_2 aa1 dd1
cmark_b  A
cmark_b  B
cmark_b  C_1
cmark_t  C
cmark_t  A_1
cmark_lt  B_1
cmark_b  C_2
cmark_b  B_2
cmark_b  A_2
cmark_t  D
cmark_b  D_2
cmark_r  D_1
drawsegment  A B
drawsegment  B C
drawsegment  C D
drawsegment  A D
drawsegment  C C_1
drawsegment  A A_1
drawsegment  B B_1
drawsegment  D D_1



Nome Utilizador: 
point  A 20 20
point  B 90 20
point  C 110 80
line  ab A B
line  bc B C
parallel  ad A bc
parallel  cd C ab
intersec  D ad cd
towards  C_1 A B 0.333333333
towards  D_1 B C 0.333333333
towards  A_1 C D 0.333333333
towards  B_1 D A 0.333333333
line  aa1 A A_1
line  bb1 B B_1
line  cc1 C C_1
line  dd1 D D_1
intersec  A_2 aa1 bb1
intersec  B_2 bb1 cc1
intersec  C_2 dd1 cc1
intersec  D_2 aa1 dd1
prove { equal { signed_area4 A B C D } { mult { 13 } { signed_area4 A_2 B_2 C_2 D_2 } } }


Nome do Teorema: chou88


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Two doubly perspective triangles are in fact
 triply perspective.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  130 100
point  O 20 20
point  A 70 20
point  B 40.40 60.70
point  C 73.20 52.50
point  O_1 57.10 64.00
line  co1 C O_1
line  ao A O
line  ao1 A O_1
line  bo B O
line  bo1 B O_1
line  co C O
intersec  A_1 co1 ao
intersec  B_1 ao1 bo
intersec  C_1 bo1 co
line  ba1 B A_1
line  ac1 A C_1
intersec  O_2 ba1 ac1
cmark_b  O
cmark_b  A
cmark_l  B
cmark_r  B_1
cmark_lt  O_1
cmark_b  C
cmark_l  O_2
cmark_rb  C_1
cmark_b  A_1
drawsegment  O B_1
drawsegment  O A_1
drawsegment  O C_1
drawsegment  A B_1
drawsegment  B C_1
drawsegment  A C_1
drawsegment  B A_1
drawsegment  O_1 A_1
line  b1co2 B_1 C
drawdashline  b1co2



Referência BiBTeX: 
point  O 20 20
point  A 70 20
point  B 40.40 60.70
point  C 73.20 52.50
point  O_1 57.10 64.00
line  co1 C O_1
line  ao A O
line  ao1 A O_1
line  bo B O
line  bo1 B O_1
line  co C O
intersec  A_1 co1 ao
intersec  B_1 ao1 bo
intersec  C_1 bo1 co
line  ba1 B A_1
line  ac1 A C_1
intersec  O_2 ba1 ac1
prove { collinear C B_1 O_2 }


Nome Utilizador: chou88


Nome do Teorema: gpreda1


Teorema: 
 Example 69 


Descrição Breve: 


Observações: 


Código GCLC: 
dim  130 100
point  A 20 20
point  B 90 20
point  C 110 70
line  ab A B
line  bc B C
parallel  cd C ab
parallel  ad A bc
intersec  D cd ad
midpoint  A_1 A B
midpoint  B_1 B C
midpoint  C_1 C D
midpoint  D_1 D A
line  ac A C
line  bd B D
intersec  O ac bd
cmark_b  A
cmark_b  B
cmark_b  A_1
cmark_t  C
cmark_t  D
cmark_t  C_1
cmark_r  B_1
cmark_l  D_1
cmark_b  O
drawsegment  A B
drawsegment  B C
drawsegment  C D
drawsegment  D A
drawsegment  A_1 B_1
drawsegment  B_1 C_1
drawsegment  C_1 D_1
drawsegment  D_1 A_1
drawdashsegment  A C
drawdashsegment  B D
drawdashsegment  A_1 C_1
drawdashsegment  B_1 D_1



Código GCLCprover: 
point  A 20 20
point  B 90 20
point  C 110 70
line  ab A B
line  bc B C
parallel  cd C ab
parallel  ad A bc
intersec  D cd ad
midpoint  A_1 A B
midpoint  B_1 B C
midpoint  C_1 C D
midpoint  D_1 D A
line  ac A C
line  bd B D
intersec  O ac bd
prove { collinear A_1 C_1 O }
prove { collinear D_1 B_1 O }


Referência BiBTeX: chou88


Nome Utilizador: gpreda1


Nome do Teorema: 
 Harmonic Points 2



Teorema: \begin{geothm}
 Given four points $A$, $B$, $C$ and $D$,
 points of intersections are formed: $L = AB \cap CD$, 
 $K = AC \cap BC$, $F = KL \cap BD$ and $G = KL \cap AC$.
 Show that segment $KL$ is harmonically divided by
 the points $F$ and $G$, ie:
 $\frac{\overline{LK}}{\overline{KF}} = \frac{\overline{GL}}{\overline{GF}}$.
\end{geothm}

Descrição Breve: 


Observações: 
dim  130 50
point  A 30 32
point  B 40 40
point  C 48 28
point  D 40 25
cmark_lt  A
cmark_t  B
cmark_rt  C
cmark_lb  D
line  AB A B
line  CD C D
intersec  L AB CD
cmark_b  L
line  AD A D
line  BC B C
intersec  K AD BC
cmark_b  K
line  BD B D
line  KL K L
intersec  F BD KL
cmark_b  F
line  AC A C
intersec  G AC KL
cmark_b  G
drawsegment  B L
drawsegment  C L
drawsegment  A K
drawsegment  B K
drawsegment  B F
drawsegment  A G
line  lkfg L K
drawdashline  lkfg



Código GCLC: 
point  A 30 32
point  B 40 40
point  C 48 28
point  D 40 25
line  AB A B
line  CD C D
intersec  L AB CD
line  AD A D
line  BC B C
intersec  K AD BC
line  BD B D
line  KL K L
intersec  F BD KL
line  AC A C
intersec  G AC KL
line  lkfg L K
prove { harmonic L K F G }



Código GCLCprover: 


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Simson s Theorem



Nome do Teorema: \begin{geothm}
 Let $D$ be a point on the circumscribed
 cirle $(O)$ of triangle $ABC$. From $D$, three
 perpendiculars are drawn to the three sides $BC$,
 $CA$ and $AB$ of triangle $ABC$. Let $E$, $F$ and $G$
 be the three feet respectively. Show that $E$, $F$ and $G$
 are collinear.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  100 100
point  A 20 50
point  O 55 50
circle  k O A
point  B 79.50 25.40
point  C 49.50 84.50
point  D 30.80 24.70
line  ab A B
line  bc B C
line  ac A C
foot  E D bc
foot  F D ac
foot  G D ab
cmark_l  A
cmark_b  B
cmark_t  C
cmark_b  D
cmark_t  O
cmark_t  E
cmark_t  F
cmark_t  G
drawsegment  A B
drawsegment  F C
drawsegment  C B
drawcircle  k
drawdashsegment  E D
drawdashsegment  F D
drawdashsegment  G D
line  efg E F
drawdashline  efg



Observações: 
point  A 20 50
point  O 55 50
circle  k O A
oncircle  B  O A
oncircle  C  O A
oncircle  D  O A
line  ab A B
line  bc B C
line  ac A C
foot  E D bc
foot  F D ac
foot  G D ab
line  efg E F
prove { collinear E F G }


Código GCLC: 


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Example 77 


Nome Utilizador: \begin{geothm}
 The cross ratio of four points on a circles
 with respect to any points on the circle
 is constant.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  A 30 30
point  B 100 30
point  C 90 80
med  m1 A B
med  m2 B C
intersec  O m1 m2
circle  k O A
oncircle  D  O A
oncircle  P  O A
oncircle  Q  O A
line  pd  P  D 
line  pc  P  C
line  qd  Q  D 
line  qc  Q  C
line  ab A B
intersec  F  pd  ab
intersec  G  pc  ab
intersec  F_1  qd  ab
intersec  G_1  qc  ab
prove { equal { mult { sratio G A G B } { sratio F B F A } } { mult { sratio G_1 A G_1 B } { sratio F_1 B F_1 A } } }


Observações: chou88


Código GCLC: gpreda1


Código GCLCprover: 
 Example 82 


Referência BiBTeX: \begin{geothm}
 In triangle $ABC$, let $F$ the midpoint of the
 side $BC$, $D$ and $E$ the feet of the
 altitudes on $AB$ and $BC$, respectively.
 $FG$ is perpendicular to $DE$ at $G$. Show that
 $G$ is the midpoint of $DE$.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  110 100
point  A 20 20
point  B 90 20
point  C 70 70
midpoint  F A B
line  ab A B
line  ac A C
line  bc B C
foot  P A bc
foot  Q B ac
line  pq P Q
foot  G F pq
cmark_b  A
cmark_b  B
cmark_b  F
cmark_t  C
cmark_r  P
cmark_l  Q
cmark_t  G
drawsegment  A B
drawsegment  C B
drawsegment  A C
drawsegment  A P
drawsegment  Q B
drawsegment  P Q
drawsegment  F G



Descrição Breve: chou88


Observações: gpreda1


Código GCLC: 
 Example 85 


Código GCLCprover: \begin{geothm}
 Let $P$ and $Q$ be two points on side $BC$
 and $AD$ of a parallelogram such that
 $PQ \parallel AB$; $M = AP \cap BQ$,
 $N = DP \cap QC$. Show that $MN \parallel AD$
 and $MN = AD/2$.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  130 100
point  A 20 20
point  B 80 20
point  C 110 70
line  ab A B
line  bc B C
parallel  ad A bc
parallel  cd C ab
intersec  D ad cd
point  P 85.45 29.09
parallel  pq P ab
intersec  Q pq ad
line  pd P D
line  qc Q C
intersec  N pd qc
line  pa P A
line  qb Q B
intersec  M pa qb
cmark_b  M
cmark_b  A
cmark_t  N
cmark_b  B
cmark_t  C
cmark_t  D
cmark_r  P
cmark_l  Q
drawsegment  P D
drawsegment  P A
drawsegment  Q C
drawsegment  B Q
drawsegment  P Q
drawsegment  A B
drawsegment  A D
drawsegment  C B
drawsegment  C D
drawdashsegment  M N



Teorema: 
point  A 20 20
point  B 80 20
point  C 110 70
line  ab A B
line  bc B C
parallel  ad A bc
parallel  cd C ab
intersec  D ad cd
online  P  B C
parallel  pq  P  ab
intersec  Q  pq  ad
line  pd  P  D
line  qc  Q  C
intersec  N  pd  qc 
line  pa  P  A
line  qb  Q  B
intersec  M  pa  qb 
prove { equal { sratio A D M N } { 2 } }


Descrição Breve: gpreda1


Observações: 
 Example 88 


Código GCLC: \begin{geothm}
 Show that the sum of distances of a point
 on the base of an isosceles triangle to its
 two sides is equal to the altitudes on the
 sides.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  110 80
point  A 20 20
point  B 90 20
point  C 33.17 60.88
point  D 23.73 31.58
line  ab A B
line  bc B C
foot  E D ab
foot  F D bc
foot  H C ab
cmark_b  A
cmark_b  B
cmark_t  C
cmark_l  D
cmark_b  H
cmark_b  E
cmark_rt  F
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C H
drawsegment  D E
drawsegment  D F



Nome do Teorema: 
point  A 20 20
point  B 90 20
oncircle  C  B A
online  D  A C
line  ab A B
line  bc  B C 
foot  E  D  ab
foot  F  D  bc 
foot  H  C  ab
prove { alg_sum_zero3 { segment D E } { segment D F } { segment C H } }


Teorema: chou88


Descrição Breve: 
 Leon Anne s Theorem



Observações: \begin{geothm}
 Given quadrilateral $ABCD$, let $M_1$ and $M_2$ be
 midpoints of diagonals $AC$ and $BD$, and let $O$
 be point on line $M_1M_2$. Show that the sum of
 areas of triangles $ABO$ and $CDO$ equals the half of the
 area of quadrilateral $ABCD$.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  110 110
point  A 20 20
point  B 90 20
point  C 75 85 
point  D 40 65
midpoint  M_1 A C
midpoint  M_2 B D
line  m12 M_1 M_2
point  O 31.05 61.90
line  ac A C
line  bd B D
drawline  ac
drawline  bd
drawdashline  m12
cmark_b  M_1
cmark_b  M_2
cmark_t  O
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  D
drawsegment  A B
drawsegment  C B
drawsegment  C D
drawsegment  A D



Nome Utilizador: 
point  A 20 20
point  B 90 20
point  C 75 85 
point  D 40 65
midpoint  M_1 A C
midpoint  M_2 B D
line  m12 M_1 M_2
online  O  M_1 M_2
line  ac A C
line  bd B D
prove { equal { mult { 2 } { sum { signed_area3 A B O } { signed_area3 C D O } } } { signed_area4 A B C D } } }


Nome do Teorema: 


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Of the three lines joining the vertices
 of an equilateral triangle to a point
 on its circumcircle, one is equal to the
 sum of the other two.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  110 100
point  A 20 30
point  B 90 30
circle  k A B
med  mc A B
intersec2  C C1 k mc
med  mb A C
intersec  O mc mb
circle  k1 O A
point  P 75.80 84.86
cmark_lb  A
cmark_rb  B
cmark_rt  P
cmark_b  O
cmark_t  C
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawdashsegment  P A
drawdashsegment  P B
drawdashsegment  P C
drawcircle  k1



Referência BiBTeX: 
point  A 20 30
point  B 90 30
circle  k A B
med  mc A B
intersec2  C C1 k mc
med  mb A C
intersec  O mc mb
circle  k1 O A
oncircle  P  O A
prove { alg_sum_zero3 { segment P A } { segment P B } { segment P C } }


Nome Utilizador: chou88


Nome do Teorema: gpreda1


Teorema: 
 Example 120 


Descrição Breve: 


Observações: 


Código GCLC: 
dim  110 100
point  A 50 50
point  B 90 50
point  C 75 80
point  X 82 66
line  ab A B
line  bc B C
line  ac A C
line  ax A X
parallel  cz C ax
parallel  by B ax
intersec  Y by ac
intersec  Z cz ab
cmark_t  Z
cmark_b  Y
cmark_rt  X      
cmark_b  A
cmark_b  B
cmark_t  C
drawsegment  Z B
drawsegment  Y C
drawsegment  Z X
drawsegment  A X
drawsegment  Y Z
drawsegment  B C
drawsegment  C Z
drawsegment  Y B
drawsegment  X Y



Código GCLCprover: 
point  A 50 50
point  B 90 50
point  C 75 80
online  X  B C
line  ab A B
line  bc B C
line  ac A C
line  ax  A X 
parallel  cz  C ax 
parallel  by  B ax 
intersec  Y  by  ac
intersec  Z  cz  ab
prove { equal { mult { 2 } { signed_area3 B A C } } { signed_area3 X Y Z } }



Referência BiBTeX: chou88


Nome Utilizador: gpreda1


Nome do Teorema: 
 Example 37 


Teorema: \begin{geothm}
 Let $PT$ and $PB$ be two tangents to a circle,
 $AB$ the diameter through $B$, and $TH$
 the perpendicular from $T$ to $AB$. Then
 $AP$ bisects $TH$.
\end{geothm}

Descrição Breve: 


Observações: 
dim  130 110
point  A 30 50
point  B 90 50 
midpoint  O A B
circle  k O A
point  T 69.10 78.59
line  ab A B
perp  bp B ab
line  ot O T
perp  tp T ot
intersec  P bp tp
foot  H T ab
line  th T H
line  ap A P
intersec  I th ap
cmark_lb  A
cmark_b  H
cmark_b  O
cmark_rb  B
cmark_r  P
cmark_t  T
cmark_rb  I
drawsegment  A B
drawsegment  A P
drawsegment  T H
drawsegment  T O
drawsegment  P B
drawsegment  P T
drawcircle  k



Código GCLC: 
point  A 30 50
point  B 90 50 
midpoint  O A B
circle  k O A
oncircle  T  O A
line  ab A B
perp  bp B ab
line  ot O T
perp  tp T ot
intersec  P bp tp
foot  H T ab
line  th T H
line  ap A P
intersec  I th ap
prove { equal { sratio T I I H } { 1 } }


Código GCLCprover: chou88


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Example 42 


Nome do Teorema: \begin{geothm}
 If a quadrangle $ABCD$ has its opposite sides $AD$
 and $BC$ (extended) meeting at $W$, while $X$ and $Y$
 are the midpoints of the diagonals $AC$ and $BD$,
 then $(WXY) = 1/4(ABCD)$.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  110 90
point  A 20 20
point  B 60 20
point  C 50 50
point  D 35 65
line  ab A B
line  cd C D
intersec  W ab cd
midpoint  X A C
midpoint  Y B D
cmark_b  X
cmark_b  Y
cmark_b  W
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  D
drawsegment  A W
drawsegment  C B
drawsegment  D W
drawsegment  A D
drawdashsegment  B D
drawdashsegment  C A
drawsegment  X Y
drawsegment  X W
drawsegment  Y W



Observações: 
point  A 20 20
point  B 60 20
point  C 50 50
point  D 35 65
line  ab A B
line  cd C D
intersec  W ab cd
midpoint  X A C
midpoint  Y B D
prove { equal { mult { 4 } { signed_area3 X W Y } } { signed_area4 A B C D } }



Código GCLC: chou88


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Desargues  theorem



Nome Utilizador: \begin{geothm}
 If lines $A_1B_1$, $A_2B_2$ and $A_3B_3$ are
 concurrent, show that intersections
 $A_1A_2 \cap B_1B_2$, $A_1A_3 \cap B_1B_3$
 and $A_2A_3 \cap B_2B_3$ are collinear.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  S 20 30
point  A_2 36.60 30
point  B_2 53.00 30
point  A_1 27.20 24.70
online  B_1  S A_1
point  A_3 42.20 41.10 
online  B_3  S A_3
line  a12 A_1 A_2
line  a13 A_1 A_3
line  a23 A_2 A_3
line  b12 B_1 B_2
line  b13 B_1 B_3
line  b23 B_2 B_3
intersec  P a12 b12
intersec  Q a13 b13
intersec  R a23 b23
prove { collinear P Q R }


Observações: 


Código GCLC: gpreda1


Código GCLCprover: 
  Six Points Circle



Referência BiBTeX: \begin{geothm}
 Given triangle $ABC$, let $A_1$, $B_1$ and $C_1$
 endpoints of triangle altitudes. From each of
 points $A_1$, $B_1$ and $C_1$ feet are drawn upon
 triangle sides, those are: $A_2$, $A_3$, $B_2$,
 $B_3$, $C_2$ and $C_3$. Show that these six
 points lie on a circle.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  130 100
point  A 20 20
point  B 90 20
point  C 70 70
line  ac A C
line  ab A B
line  bc B C
foot  A_1 A bc
foot  B_1 B ac
foot  C_1 C ab
foot  C_2 C_1 bc
foot  C_3 C_1 ac
foot  A_2 A_1 ac
foot  A_3 A_1 ab
foot  B_2 B_1 bc
foot  B_3 B_1 ab
med  p B_3 A_3
med  q B_3 C_3
intersec  O p q
circle  k O B_3
cmark_b  A
cmark_b  B
cmark_t  C
cmark_r  A_1
cmark_l  B_1
cmark_b  C_1
cmark_b  A_3
cmark_b  B_3
cmark_l  C_3
cmark_lt  A_2
cmark_r  C_2
cmark_rt  B_2
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawdashsegment  A A_1
drawdashsegment  B B_1
drawdashsegment  C C_1
drawdashsegment  C_1 C_2
drawdashsegment  C_1 C_3
drawdashsegment  A_1 A_2
drawdashsegment  A_1 A_3
drawdashsegment  B_1 B_2
drawdashsegment  B_1 B_3
drawcircle  k



Descrição Breve: 


Observações: gpreda1


Código GCLC: 
 Quadrilateral Inscribed in Circle



Código GCLCprover: \begin{geothm}
 If quadrilateral $ABCD$ is inscribed in circle,
 show that angles $\angle ACB$ and $\angle ADB$
 are equal.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  110 90
point  A 40 20
point  B 70 20
point  X 56 45
line  ax A X
med  m A B
intersec  O ax m
circle  k O A
drawcircle  k
point  C 33.76 61.42
point  D 69.32 67.30
drawsegment  C A
drawsegment  C B
drawsegment  D A
drawsegment  D B
cmark_lt  C
cmark_rt  D
cmark_lb  A
cmark_rb  B
drawsegment  A B



Teorema: 
point  A 40 20
point  B 70 20
point  X 56 45
line  ax A X
med  m A B
intersec  O ax m
circle  k O A
oncircle  C  O A
oncircle  D  O A 
prove { equal { angle A C B } { angle A D B } }



Descrição Breve: gpreda1


Observações: 
 Example 30 


Código GCLC: \begin{geothm}
 Let $H$ and $O$ be the orthocenter
 and circumcenter of triangle $ABC$.
 Show
 $\angle HAC = | \angle B - \angle C |$.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  130 110
point  C 20 30
point  B 110 30
point  A 90 95
line  ab A B
line  ac A C
line  bc B C
foot  D A bc
foot  B_1 B ac
line  ha A D
line  hb B B_1
intersec  H ha hb
med  ma A B
med  mb A C
intersec  O ma mb
circle  k O A
cmark_lb  C
cmark_rb  B
cmark_rt  A
cmark_r  H
cmark_l  O
cmark_b  D
drawsegment  A B
drawsegment  A C
drawsegment  B C
drawsegment  A D
drawcircle  k
drawsegment  O A
drawsegment  H A



Nome do Teorema: 
point  C 20 30
point  B 110 30
point  A 90 95
line  ab A B
line  ac A C
line  bc B C
foot  D A bc
med  ma A B
med  mb A C
intersec  O ma mb
prove { equal { sum { angle O A D } { mult { mult { mult { angle A B C } { angle O A D } } { angle A C B } } { -1 } } } { sum { angle A B C } { angle A C B } } }


Teorema: chou88


Descrição Breve: 
 Pascal s Theorem



Observações: \begin{geothm}
 Hexagon $ABCDEF$ is incsibed in a circle with
 center $O$. Show that three points at which
 pairs of opposite sides meet, lie on a
 straight line.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  100 100
point  A 20.50 51.40
point  O 50 50
circle  k O A
point  B 60.40 77.60
point  C 72.30 69.20
point  D 21.40 42.10
point  E 51.30 20.60
point  F 78.20 41.90
line  ab A B
line  df D F
line  bc B C
line  fe F E
line  cd C D
line  ea E A
drawdashline  ab
drawdashline  df
drawdashline  bc
drawdashline  fe
drawdashline  cd
drawdashline  ea
intersec  P ab df
intersec  Q bc fe
intersec  S cd ea
cmark_lt  A
cmark_t  O
cmark_t  B
cmark_t  C
cmark_lb  D
cmark_t  E
cmark_lt  F
cmark_t  P
cmark_t  Q
cmark_t  S
drawcircle  k
line  pqs P Q
drawline  pqs



Nome Utilizador: 
point  A 20.50 51.40
point  O 50 50
circle  k O A
oncircle  B  O A
oncircle  C  O A
oncircle  D  O A
oncircle  E  O A
oncircle  F  O A
line  ab A B
line  df D F
line  bc B C
line  fe F E
line  cd C D
line  ea E A
intersec  P ab df
intersec  Q bc fe
intersec  S cd ea
line  pqs P Q
prove { collinear P Q S }


Nome do Teorema: 


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Let $ABC$ be a triangle with $\angle B = 2\angle C$,
 D the foot of the altitude on $CB$ and $M$ the
 midpoint of $B$ and $C$. Show that $AB = 2DM$.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  110 80
point  C 20 20
point  B 90 20
line  bc B C
med  m B C
point  A_1 80 60
line  ab A_1 B
intersec  F ab m
bis  ca B C F
intersec  A ca ab
intersec  M bc m
foot  D A bc
cmark_b  C
cmark_b  B
cmark_r  A
cmark_b  M
cmark_b  D
drawsegment  B C
drawsegment  A C
drawsegment  B A
drawsegment  D A
drawsegment  M A



Referência BiBTeX: 
point  C 20 20
point  B 90 20
line  bc B C
med  m B C
point  A_1 80 60
line  ab A_1 B
intersec  F ab m
bis  ca B C F
intersec  A ca ab
intersec  M bc m
foot  D A bc
prove { equal { mult { 4 } { segment D M } } { segment A B } }


Nome Utilizador: chou88


Nome do Teorema: gpreda1


Teorema: 
 Example 20 from \cite{geothms}



Descrição Breve: 


Observações: 


Código GCLC: 
dim  110 100
point  A 20 20
point  C 65 80
point  T 55 45
towards  A_1 A T 1.5
towards  C_1 C T 1.5
line  bc C A_1
line  ab A C_1
intersec  B ab bc
point  X 95 45
line  gx T X
foot  D A gx
foot  E B gx
foot  F C gx
cmark_t  D
cmark_t  E
cmark_b  F
cmark_lt  T
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  X
drawline  gx
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  A A_1
drawsegment  C C_1
drawdashsegment  A D
drawdashsegment  B E
drawdashsegment  C F



Código GCLCprover: 
point  A 20 20
point  C 65 80
point  T 55 45
towards  A_1 A T 1.5
towards  C_1 C T 1.5
line  bc C A_1
line  ab A C_1
intersec  B ab bc
point  X 95 45
line  gx T X
foot  D A gx
foot  E B gx
foot  F C gx
prove { equal { sum { sum { sratio A D A D } { sratio B E A D } } { sratio C F A D } } { 0 } }


Referência BiBTeX: 


Nome Utilizador: gpreda1


Nome do Teorema: 
 Example 22 from \cite{geothms}



Teorema: \begin{geothm}
 Given a parallelogram $ABCD$, a point $N$,
 obtained by the intersection of a line parallel
 to $BD$ passing through $C$, and a line perpendicular
 to $BD$ passing through $A$, then the point $Q$,
 which is given by the intersection of $BN$ and $CD$,
 is the midpoint of $PC$, where $P$ is the intersection
 of $CD$ and $AN$,
\end{geothm}

Descrição Breve: 


Observações: 
dim  110 90
point  A 20 20
point  B 70 20
point  C 85 60
line  ab A B
line  bc B C
parallel  ad A bc
parallel  cd C ab
intersec  D ad cd
line  bd B D
parallel  cn C bd
perp  an A bd
intersec  N cn an
line  bn B N
intersec  P cd an
intersec  Q bn cd
drawline  cn
drawline  an
drawsegment  A N
drawsegment  B N
drawsegment  B D
cmark_lt  P
cmark_lb  Q
cmark_t  N
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  D
drawsegment  A B
drawsegment  C B
drawsegment  C D
drawsegment  A D



Código GCLC: 
point  A 20 20
point  B 70 20
point  C 85 60
line  ab A B
line  bc B C
parallel  ad A bc
parallel  cd C ab
intersec  D ad cd
line  bd B D
parallel  cn C bd
perp  an A bd
intersec  N cn an
line  bn B N
intersec  P cd an
intersec  Q bn cd
prove { equal { sratio P Q Q C } { 1 } }


Código GCLCprover: 


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Nobbs  Theorem



Nome do Teorema: \begin{geothm}
 Assume $S_1$, $S_2$, $S_3$ are the points of contact 
 of the incircle of $ABC$: $S_1$ on side $AC$, etc.
 Denote the intersection of $BC$ and $S_1S_3$ as $P$,
 that of $AC$ and $S_2S_3$ as $Q$, and let $R$ 
 be the intersection of $AB$ 
 and $S_1S_2$. Then points $P$, $Q$, and $R$ are collinear.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  130 130
point  X 20 20
point  Y 25 20
point  S_3 65 20
point  O 65 32
line  ab X Y
circle  k O S_3
point  S_1 75.90 36.70
point  S_2 60.10 42.90
line  os2 O S_2
line  os1 O S_1
perp  as1 S_1 os1
perp  bs2 S_2 os2
intersec  A ab as1
intersec  B ab bs2
intersec  C as1 bs2
line  s13 S_1 S_3
intersec  P s13 bs2
line  s23 S_2 S_3
intersec  Q s23 as1
line  s12 S_1 S_2
intersec  R s12 ab
line  pqr P Q
drawcircle  k
drawline  pqr
drawdashline  s23
drawdashline  s13
drawdashline  s12
drawdashsegment  C P
drawdashsegment  C Q
drawdashsegment  A R
cmark_b  R
cmark_rt  Q
cmark_r  P
cmark_t  C
cmark_b  A
cmark_b  B
cmark_l  S_1
cmark_b  S_2
cmark_rb  S_3
drawsegment  A C
drawsegment  A B
drawsegment  B C



Observações: 
point  X 20 20
point  Y 25 20
point  S_3 65 20
point  O 65 32
line  ab X Y
circle  k O S_3
oncircle  S_1  O S_3
oncircle  S_2  O S_3
line  os2 O S_2
line  os1 O S_1
perp  as1 S_1 os1
perp  bs2 S_2 os2
intersec  A ab as1
intersec  B ab bs2
intersec  C as1 bs2
line  s13 S_1 S_3
intersec  P s13 bs2
line  s23 S_2 S_3
intersec  Q s23 as1
line  s12 S_1 S_2
intersec  R s12 ab
line  pqr P Q
prove { collinear P Q R }


Código GCLC: 


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Menelaus s Theorem



Nome Utilizador: \begin{geothm}
 Line $p$ intersects lines $AB$, $AC$ and $BC$
 at points $F$, $E$ and $D$ respectively. Show that:
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  A 30 20
point  B 60 20
point  C 40 50
line  c A B
online  D  B C 
online  E  A C 
line  p D E 
intersec  F c p
prove { equal { mult { mult { sratio A F F B } { sratio B D D C } } { sratio C E E A } } -1 }



Observações: 


Código GCLC: gpreda1


Código GCLCprover: 
 Gauss  line



Referência BiBTeX: \begin{geothm}
 The midpoints of the three diagonals
 of a complete quadrilateral are collinear.
 This line is called Gauss line.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  90 80
point  A_0 30 10
point  A_1 70 10
point  A_2 55 40
point  A_3 35 25
line  a12 A_1 A_2
line  a03 A_0 A_3
line  a01 A_0 A_1
line  a23 A_2 A_3
intersec  X a12 a03
intersec  Y a01 a23
midpoint  M_1 A_1 A_3
midpoint  M_2 A_0 A_2
midpoint  M_3 X Y 
cmark_b  A_0
cmark_b  A_1
cmark_rt  A_2
cmark_l  A_3
cmark_t  X
cmark_b  Y
cmark_b  M_1
cmark_r  M_2
cmark_rt  M_3
drawsegment  A_0 A_1
drawsegment  A_0 A_2
drawsegment  A_0 A_3
drawsegment  A_1 A_2
drawsegment  A_1 A_3
drawsegment  A_2 A_3
drawdashsegment  A_1 X
drawdashsegment  A_3 X
drawdashsegment  A_0 Y
drawdashsegment  A_2 Y
drawline  M_1 M_2



Descrição Breve: 


Observações: gpreda1


Código GCLC: 
 Example 25 


Código GCLCprover: \begin{geothm}
 A tranversal curs the sides $BC$, $CA$, $AB$
 of triangle $ABC$ in $D$, $E$, $F$. $P$, 
 $Q$, $R$ are the midpoints of $EF$, $FD$,
 $DE$ and $AP$, $BQ$, $CR$ intersec $BC$, $CA$,
 $AB$ in $X$, $Y$, $Z$. Show that $X$, $Y$, $Z$ are collinear.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  150 110
point  A 20 20
point  B 70 20
point  C 60 70
line  ab A B
line  bc B C
line  ac A C
point  F 63 20
point  D 66.06 39.70
line  fd F D
intersec  E fd ac
midpoint  P E F
midpoint  Q F D
midpoint  R D E
line  ap A P
line  bq B Q
line  cr C R
intersec  X ap bc
intersec  Y bq ac
intersec  Z cr ab
line  xyz X Y
drawdashline  xyz
cmark_lb  X
cmark_t  Y
cmark_b  Z
cmark_r  P
cmark_l  Q
cmark_r  R
cmark_t  E
cmark_b  F
cmark_r  D
cmark_b  A
cmark_b  B
cmark_t  C
drawsegment  A P
drawsegment  B Y
drawsegment  C Z
drawsegment  A Z
drawsegment  C B
drawsegment  A E
drawsegment  F E



Teorema: 
point  A 20 20
point  B 70 20
point  C 60 70
line  ab A B
line  bc B C
line  ac A C
online  F  A B
online  D  B C
line  fd F D
intersec  E fd ac
midpoint  P E F
midpoint  Q F D
midpoint  R D E
line  ap A P
line  bq B Q
line  cr C R
intersec  X ap bc
intersec  Y bq ac
intersec  Z cr ab
line  xyz X Y
prove { collinear X Y Z }


Descrição Breve: gpreda1


Observações: 
 Example 28 


Código GCLC: \begin{geothm}
 In a hexagon $AC_1BA_1CB_1$, $BB_1$, $C_1A$,
 $A_1C$ are concurrent and $CC_1$, $A_1B$, $B_1A$
 are concurrent. Prove that $AA_1$, $B_1C$, $C_1B$
 are also concurrent.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  110 90
point  A 30 20
point  C_1 58 20
point  B 20 37
point  A_1 37 50
point  C 60 35
line  ac1 A C_1
line  a1c A_1 C
intersec  O ac1 a1c
line  ba1 B A_1
line  cc1 C C_1
intersec  H ba1 cc1
line  ah A H
line  bo B O
intersec  B_1 bo ah
line  cb1 C B_1
line  bc1 B C_1
intersec  I cb1 bc1
cmark_rb  B_1
cmark_t  H
cmark_rb  A
cmark_b  C_1
cmark_lb  B
cmark_lt  A_1
cmark_rt  C
cmark_b  O
drawsegment  B H
drawsegment  A H
drawsegment  C_1 H
drawsegment  A O
drawsegment  B O
drawsegment  A_1 O
drawsegment  A B
drawsegment  B C
drawdashline  C B_1
drawdashline  B C_1
drawdashline  A A_1



Nome do Teorema: 
point  A 30 20
point  C_1 58 20
point  B 20 37
point  A_1 37 50
point  C 60 35
line  ac1 A C_1
line  a1c A_1 C
intersec  O ac1 a1c
line  ba1 B A_1
line  cc1 C C_1
intersec  H ba1 cc1
line  ah A H
line  bo B O
intersec  B_1 bo ah
line  cb1 C B_1
line  bc1 B C_1
intersec  I cb1 bc1
prove { collinear A A_1 I }


Teorema: chou88


Descrição Breve: 
 Example 29 


Observações: \begin{geothm}
 In a triangle $ABC$, let $p$ and $q$ be the radii of two
 circles through $A$, touching side $BC$ at $B$ and $C$,
 respectively. Then $pq = R^2$.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  90 90
point  B 20 20
point  C 70 20
point  A 55 60
med  ma B C
med  mb A C
intersec  O ma mb
line  bc B C
perp  p B bc
med  mc A B
intersec  P p mc
perp  q C bc
intersec  Q mb q
circle  k1 Q A
circle  k2 P A
drawcircle  k1
drawcircle  k2
cmark_t  P
cmark_t  Q
cmark_b  B
cmark_b  C
cmark_t  A
cmark_t  O
drawsegment  P B
drawsegment  P A
drawsegment  Q C
drawsegment  Q A
drawdashsegment  O B
drawdashsegment  O C
drawsegment  A B
drawsegment  C B
drawsegment  A C



Nome Utilizador: 
point  B 20 20
point  C 70 20
point  A 55 60
med  ma B C
med  mb A C
intersec  O ma mb
line  bc B C
perp  p B bc
med  mc A B
intersec  P p mc
perp  q C bc
intersec  Q mb q
prove { equal { mult { segment Q C } { segment P B } } { mult { segment O B } { segment O B } } }


Nome do Teorema: chou88


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 $OH^2 = 9R^2 - a^2 - b^2 - c^2$.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  90 90
point  A 20 20
point  B 70 20
point  C 55 70
med  ma B C
med  mc A B
intersec  O ma mc
line  ab A B
perp  ch C ab
line  ac A C
perp  bh B ac
intersec  H ch bh
intersec  D ch ab
circle  k O A
drawcircle  k
cmark_b  O
cmark_l  H
cmark_lb  A
cmark_rb  B
cmark_t  C
drawsegment  C D
drawsegment  C O
drawsegment  A O
drawsegment  A B
drawsegment  A C
drawsegment  C B



Referência BiBTeX: 
point  A 20 20
point  B 70 20
point  C 55 70
med  ma B C
med  mc A B
intersec  O ma mc
line  ab A B
perp  ch C ab
line  ac A C
perp  bh B ac
intersec  H ch bh
prove { equal { mult { 9 } { segment O A } } { sum { segment O H } { sum { segment A B } { sum { segment A C } { segment B C } } } } }


Nome Utilizador: chou88


Nome do Teorema: gpreda1


Teorema: 
 Viviani s Theorem



Descrição Breve: 


Observações: 


Código GCLC: 
dim  110 100
point  A 20 20
point  B 90 20
point  P 65 40
circle  k A B
med  p A B
intersec2  C C1 k p
line  ab A B
intersec  D ab p
line  bc B C
line  ac A C
foot  G P ac
foot  F P bc
foot  E P ab
foot  H P p
cmark_b  D
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  P
cmark_l  H
cmark_lt  G
cmark_rt  F
cmark_b  E
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C D
drawsegment  P H
drawsegment  P G
drawsegment  P F
drawsegment  P E



Código GCLCprover: 
point  A 20 20
point  B 90 20
point  P 65 40
circle  k A B
med  p A B
intersec2  C C1 k p
line  ab A B
intersec  D ab p
line  bc B C
line  ac A C
foot  G P ac
foot  F P bc
foot  E P ab
foot  H P p
prove { alg_sum_zero3 { segment C H } { segment P F } { segment P G } }


Referência BiBTeX: 


Nome Utilizador: gpreda1


Nome do Teorema: 
 Incenter Exists Theorem



Teorema: \begin{geothm}
 Show that every triangle has incenter.
\end{geothm}

Descrição Breve: 


Observações: 
dim  110 110
point  A 20 20
point  B 90 20
point  C 70 90
bis  sa B A C
bis  sc A C B
bis  sb C B A
intersec  S sa sb
line  ab A B
intersec  C1 sc ab
cmark_rb  S
drawdashline  sa
drawdashline  sb
drawdashline  sc
cmark_b  A
cmark_b  B
cmark_lt  C
drawsegment  A B
drawsegment  C B
drawsegment  A C



Código GCLC: 
point  A 20 20
point  B 90 20
point  C 70 90
bis  sa B A C
bis  sc A C B
bis  sb C B A
intersec  S sa sb
line  ab A B
prove { equal { angle A C S } { angle S C B } }


Código GCLCprover: 


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Example 152 


Nome do Teorema: \begin{geothm}
 The feet of the four perpendiculars
 dropped from a vertex of a triangle
 upon the four bisectors of the other
 two angles are collinear.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  120 110
point  A 30 30
point  B 100 30
point  C 50 88
bis  s1 B A C
bis  s2 A B C
perp  s3 A s1
perp  s4 B s2
foot  P C s1
foot  Q C s2
foot  R C s3
foot  S C s4
cmark_b  A
cmark_rb  B
cmark_t  C
cmark_b  P
cmark_lb  Q
cmark_lb  R
cmark_rb  S
line  p P Q
drawline  p
drawdashline  s1
drawdashline  s2
drawdashline  s3
drawdashline  s4
line  a A B
drawline  a
drawsegment  A C
drawsegment  C B
drawsegment  C P
drawsegment  C Q
drawsegment  C R
drawsegment  C S



Observações: 
point  A 30 30
point  B 100 30
point  C 50 88
bis  s1 B A C
bis  s2 A B C
foot  P C s1
foot  Q C s2
line  p P Q
prove { parallel P Q A B }



Código GCLC: chou88


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Isosceles Right Triangle Problem



Nome Utilizador: \begin{geothm}
 Isosceles right triangle $ABC$ with right angle
 at point $C$ is given. Let $E$ be a point on side $AC$
 of a triangle. Let $D$ be the intersection of a line
 passing through $E$ parallel with line $AB$ and a line.
 $BC$. Let $p$ and $q$ be lines perpendicular with
 line $BE$ passing through points $D$ and $C$, and
 let $K$ and $L$ be intersections of lines $p$ and $q$
 with hypotenuse $AB$. Show that $L$ is a midpoint
 of a segment $KA$.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  C 20 20
point  A 90 20
circle  k C A
line  ac A C
perp  bc C ac
intersec2  B B1 bc k
online  E  A C
line  ab A B
line  be B E
parallel  ed E ab
intersec  D ed bc
perp  cl C be
perp  dk D be
intersec  L cl ab
intersec  K dk ab
prove { equal { sratio K L L A } { 1 } }


Observações: 


Código GCLC: gpreda1


Código GCLCprover: 
 Butterfly theorem



Referência BiBTeX: \begin{geothm}
 $P_1$, $P_2$, $P_3$ and $P_4$ are four points on
 circle $k$ with a center $O$. 
 $M$ is the intersection of $P_1P_3$ and
 $P_2P_4$. Through $M$ draw a line $l$ perpendicular
 to $OM$, meeting $P_2P_3$ at $X$ and $P_1P_4$ at $Y$.
 Show that $MX \cong MY$.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  100 100
point  P_1 32 40
point  O 62 40
circle  k O P_1
point  P_2 35.80 25.50
point  P_3 62.80 69.90
point  P_4 77.80 14.80 
line  p1p3 P_1 P_3
line  p2p4 P_2 P_4
line  p1p4 P_1 P_4
line  p2p3 P_2 P_3
intersec  M p1p3 p2p4
line  om O M
perp  l M om
intersec  X l p2p3
intersec  Y l p1p4
cmark_l  P_1
cmark_l  P_2
cmark_lt  P_3
cmark_b  P_4
cmark_b  O
cmark_lt  M
cmark_l  X
cmark_t  Y
drawcircle  k
drawdashline  p1p3
drawdashline  p2p4
drawdashline  p1p4
drawdashline  p2p3
drawsegment  X Y



Descrição Breve: 


Observações: gpreda1


Código GCLC: 
 Right Angle over Circle Diameter



Código GCLCprover: \begin{geothm}
 Prove that the angle over circle diameter is right angle.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  110 110
point  A 20 55
point  B 90 55
midpoint  O A B
circle  k O A
point  C 76.49 82.62
cmark_t  C
drawsegment  A C
drawsegment  B C
drawcircle  k
cmark_l  A
cmark_r  B
drawsegment  A B



Teorema: 
point  A 20 55
point  B 90 55
midpoint  O A B
circle  k O A
oncircle  C  O A
prove { equal { ratio 1 { angle A C B } } { 0 } }


Descrição Breve: gpreda1


Observações: 
 Ptolemy s Theorem



Código GCLC: \begin{geothm}
 Let a quadrilateral $ABCD$ be inscribed in a circle. 
 Then the sum of the products of the two pairs of 
 opposite sides equals the product of its 
 two diagonals. In other words,
 $AD \cdot BC + AB \cdot CD = AC \cdot BD$.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  110 95
point  A 20 30
point  B 90 30
point  C 60 80
med  p A B
med  q B C      
intersec  O p q
circle  k O A
point  D 30.42 71.10
cmark_lb  A
cmark_rb  B
cmark_rt  C
cmark_lt  D
cmark_b  O
drawsegment  A B
drawsegment  A D
drawsegment  A D
drawsegment  C B
drawsegment  D B
drawsegment  C D
drawsegment  C A
drawcircle  k



Nome do Teorema: 
point  A 20 30
point  B 90 30
point  C 60 80
med  p A B
med  q B C      
intersec  O p q
circle  k O A
oncircle  D  O A
prove { alg_sum_zero3 { mult { segment A B } { segment C D } } { mult { segment A D } { segment B C } } { mult { segment A C } { segment B D } } }


Teorema: 


Descrição Breve: 
 Ceva s theorem



Observações: \begin{geothm}
 There is a point $P$ inside of triangle $ABC$.
 Lines $PA$, $PB$ and $PC$ meet the triangle at three points
 $D$, $E$ and $F$. Show that:
 $\frac{\overline{AF}}{\overline{FB}} \cdot 
 \frac{\overline{BD}}{\overline{DC}} \cdot 
 \frac{\overline{CE}}{\overline{EA}} = 1$.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  100 100
point  A 60 10
point  B 30 90
point  C 80 90
point  P 55 75
line  bc B C
line  pa P A
line  ac A C
line  pb P B
line  ab A B
line  pc P C
intersec  D bc pa
intersec  E ac pb
intersec  F ab pc
drawsegment  A B
drawsegment  A C
drawsegment  B C
drawsegment  A D
drawsegment  B E
drawsegment  C F
cmark_b  A
cmark_t  B
cmark_t  C
cmark_t  D
cmark_l  F
cmark_r  E
cmark_rt  P



Nome Utilizador: 
point  A 60 10
point  B 30 90
point  C 80 90
point  P 55 75
line  bc B C
line  pa P A
line  ac A C
line  pb P B
line  ab A B
line  pc P C
intersec  D bc pa
intersec  E ac pb
intersec  F ab pc
prove { equal { mult { mult { sratio A F F B } { sratio B D D C } } { sratio C E E A } } 1 }



Nome do Teorema: 


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Given triangle $ABC$, let $p$ be line through $A$ parallel
 with $BC$, and let $q$ be line through $C$ parallel
 with $AB$. Let $p$ and $q$ intersects at $D$. 
 Let $BD$ intersect $AC$ at $E$. Show that:
 $\overline{AE} = \overline{EC}$.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  130 70
point  A 20 20
point  B 80 20
point  C 97.60 51.50
line  ab A B
line  bc B C
parallel  p A bc
parallel  q C ab
intersec  D p q
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  D
drawsegment  A B
drawsegment  C B
drawsegment  A D
drawsegment  C D
line  ac A C
line  bd B D
intersec  E ac bd
cmark_t  E
drawdashsegment  A C
drawdashsegment  B D



Referência BiBTeX: 
point  A 20 20
point  B 80 20
point  C 97.60 51.50
line  ab A B
line  bc B C
parallel  p A bc
parallel  q C ab
intersec  D p q
line  ac A C
line  bd B D
intersec  E ac bd
prove { equal { sratio A E E C } { 1 } }
prove { same_length A E E C }


Nome Utilizador: 


Nome do Teorema: gpreda1


Teorema: 
 Circumscribed Circle Exists Theorem



Descrição Breve: 


Observações: 


Código GCLC: 
dim  95 95
point  A 30 20
point  B 70 20
point  C 60 80 
med  b A C 
med  a C B 
intersec  O a b 
midpoint  M A B
drawsegment  A B
drawsegment  A C
drawsegment  B C
drawsegment  A O
drawsegment  B O
drawsegment  C O
cmark_b  A
cmark_b  B
cmark_t  C
cmark_l  O
drawcircle  O A 
drawdashsegment  O M
cmark_b  M



Código GCLCprover: 
point  A 30 20
point  B 70 20
point  C 60 80 
med  b A C 
med  a C B 
intersec  O a b 
midpoint  M A B
prove { perpendicular O M A B }



Referência BiBTeX: 


Nome Utilizador: gpreda1


Nome do Teorema: 
 Varignon s Theorem



Teorema: \begin{geothm}
 If we draw line segments between the midpoints,
 in order, of any quadrilateral, we get a parallelogram 
 which has half the area of the original 
 quadrilateral.
\end{geothm}

Descrição Breve: 


Observações: 
dim  100 100
point  A 20 20
point  B 80 20
point  C 70 70
point  D 30 50
midpoint  P A B
midpoint  Q B C
midpoint  R C D
midpoint  S D A
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  D
cmark_b  P
cmark_r  Q
cmark_t  R
cmark_l  S
drawsegment  A B
drawsegment  C B
drawsegment  C D
drawsegment  A D
drawdashsegment  P Q
drawdashsegment  R Q
drawdashsegment  R S
drawdashsegment  P S



Código GCLC: 
point  A 20 20
point  B 80 20
point  C 70 70
point  D 30 50
midpoint  P A B
midpoint  Q B C
midpoint  R C D
midpoint  S D A
prove { parallel P Q R S }
prove { same_length P Q R S }


Código GCLCprover: 


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Eight Points Circle



Nome do Teorema: \begin{geothm}
 Quadrilateral $ABCD$ is given such that diagonals
 $AC$ and $BD$ intersect at right angle. Let $A_1$,
 $B_1$, $C_1$ and $D_1$ be midpoints of quadrilateral
 sides $AB$, $BC$, $CD$ and $DA$ respectively.
 Let $A_2$, $B_2$, $C_2$ and $D_2$ be feet from points
 $A_1$, $B_1$, $C_1$ and $D_1$ upon lines $CD$, $DA$,
 $AB$ and $BC$ respectively. Show that eight points
 $A_1$, $B_1$, $C_1$, $D_1$, $A_2$, $B_2$, $C_2$ and $D_2$
 lie on a circle.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  130 110
point  A 20 35
point  C 110 35
point  B 80 20
point  D 80 75
midpoint  A_1 A B
midpoint  B_1 B C
midpoint  C_1 C D
midpoint  D_1 D A
line  cd C D
line  da D A
line  ab A B
line  bc B C
foot  A_2 A_1 cd
foot  B_2 B_1 da
foot  C_2 C_1 ab
foot  D_2 D_1 bc
midpoint  O A_1 C_1
circle  k O A_1
drawcircle  k
drawsegment  A B
drawsegment  C B
drawsegment  A D
drawsegment  A C
drawsegment  D B
drawsegment  D C
drawdashsegment  A_1 A_2
drawdashsegment  B_1 B_2
drawdashsegment  C_1 C_2
drawdashsegment  D_1 D_2
drawdashsegment  B D_2
drawdashsegment  B C_2
cmark_b  A
cmark_b  B
cmark_b  C
cmark_t  D
cmark_b  A_1
cmark_rb  B_1
cmark_rt  C_1
cmark_lt  D_1
cmark_rt  A_2
cmark_lt  B_2
cmark_b  C_2
cmark_b  D_2



Observações: 
point  A 20 35
point  C 110 35
point  B 80 20
point  D 80 75
midpoint  A_1 A B
midpoint  B_1 B C
midpoint  C_1 C D
midpoint  D_1 D A
line  cd C D
line  da D A
line  ab A B
line  bc B C
foot  A_2 A_1 cd
foot  B_2 B_1 da
foot  C_2 C_1 ab
foot  D_2 D_1 bc
midpoint  O A_1 C_1
circle  k O A_1
prove { perpendicular A_1 B_1 B_1 C_1 }
prove { perpendicular A_1 D_1 D_1 C_1 }
prove { perpendicular A_1 A_2 A_2 C_1 }



Código GCLC: 


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Example 84 


Nome Utilizador: \begin{geothm}
 A line passing through the intersection $O$ of the
 diagonals of parallelogram $ABCD$ meets the four sides
 at $E$, $F$, $G$, $H$. Show that $EF \cong GH$.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  A 30 30
point  B 110 30
point  C 90 80
line  ab A B
line  bc B C
parallel  cd C ab
parallel  ad A bc
intersec  D cd ad
online  H  B C
line  bd B D
line  ac A C
intersec  O ac bd
line  oh O H
intersec  F oh cd
intersec  G oh ab
intersec  E oh ad
prove { equal { sratio E F G H } { 1 } }


Observações: chou88


Código GCLC: gpreda1


Código GCLCprover: 
 Example 91 


Referência BiBTeX: \begin{geothm}
 Let $D$ be a point on the side $CB$ of a right triangle
 $ABC$ such that the circle ($O$) with diameter $CD$
 touches the hypotenuses $AB$ at $E$. Let
 $F = AC \cap DE$. Show that $AF \cong AE$
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  130 110
point  C 20 30
point  D 70 30
line  cd C D
midpoint  O C D
circle  k O C
point  E 53.34 53.57
line  oe O E
perp  ab E oe
perp  ac C cd
intersec  A ac ab
intersec  B ab cd
line  de D E
intersec  F de ac
cmark_lb  C
cmark_b  O
cmark_rb  D
cmark_rt  E
cmark_l  A
cmark_b  B
cmark_l  F
drawcircle  k
drawsegment  C B
drawsegment  A B
drawsegment  C F
drawsegment  O E
drawsegment  D F



Descrição Breve: chou88


Observações: gpreda1


Código GCLC: 
 Three squares



Código GCLCprover: \begin{geothm}
 Three equal squares are placed side by side,
 as it is shown on the picture. Show that:
 $\angle D_1AD + \angle D_1BD = \angle D_1CD$.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  130 90
point  A 20 20
point  B 50 20
towards  C A B 2
towards  D A B 3
line  ab A B
perp  a A ab
circle  k A B
intersec2  A_1 A_2 k a
perp  ab1 A_1 a
perp  b B ab
perp  c C ab
perp  d D ab
intersec  B_1 ab1 b
intersec  C_1 ab1 c
intersec  D_1 ab1 d
cmark_t  A_1
cmark_b  A
cmark_b  B
cmark_b  C
cmark_b  D
cmark_t  D_1
drawsegment  A D
drawsegment  A_1 D_1
drawsegment  A A_1
drawsegment  B B_1
drawsegment  C C_1
drawsegment  D D_1
drawdashsegment  D_1 A
drawdashsegment  D_1 B
drawdashsegment  D_1 C



Teorema: 
point  A 20 20
point  B 50 20
towards  C A B 2
towards  D A B 3
line  ab A B
perp  a A ab
circle  k A B
intersec2  A_1 A_2 k a
perp  ab1 A_1 a
perp  d D ab
intersec  D_1 ab1 d
prove { equal { angle D_1 C D } { sum { mult { angle D_1 A D } { mult { angle D_1 B D } { angle D_1 C D } } } { sum { angle D_1 A D } { angle D_1 B D } } } }


Descrição Breve: gpreda1


Observações: 
 Example 92 


Código GCLC: \begin{geothm}
 From the ends $D$ and $C$ of a diameter of circle $(O)$
 perpendiculars are drawn to chort $AB$. Let $E$ and $F$
 be the feet of the perpendiculars. Show that
 $OE \cong OF$.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  130 110
point  A 25 30
point  B 105 30
point  X 100 100
point  Y 92 90
line  xy X Y
med  m A B
intersec  O m xy
circle  k O A
point  Z 80 80
line  cd O Z
intersec2  D C k cd
line  ab A B
foot  E C ab
foot  F D ab
cmark_lb  A
cmark_rb  B
cmark_lt  O
cmark_lb  C
cmark_rt  D
cmark_t  E
cmark_b  F
drawcircle  k
drawdashsegment  O E
drawdashsegment  O F
drawsegment  A B
drawsegment  C D
drawsegment  C E
drawsegment  D F



Nome do Teorema: 
point  A 25 30
point  B 105 30
point  X 100 100
point  Y 92 90
line  xy X Y
med  m A B
intersec  O m xy
circle  k O A
point  Z 80 80
line  cd O Z
intersec2  D C k cd
line  ab A B
foot  E C ab
foot  F D ab
prove { equal { segment O E } { segment O F } }


Teorema: chou88


Descrição Breve: 
 Example 94 


Observações: \begin{geothm}
 Through $P$ a tangent $PE$ and a secant
 $PAB$ of circle $(O)$ are drawn. The
 bisector of angle $APE$ meets $AE$ and
 $BE$ at $C$ and $D$. Show that
 $EC \cong ED$.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  110 90
point  E 35 20
point  O 35 45
circle  k O E
point  A 17.71 63.06
point  B 58.58 36.68
line  ab A B
line  oe O E
perp  ep E oe
intersec  P ep ab
line  ae A E
line  be B E
bis  p E P A
intersec  C ae p
intersec  D be p
cmark_t  O
cmark_b  E
cmark_b  P
cmark_lt  A
cmark_r  B
cmark_lb  C
cmark_t  D
drawsegment  E P
drawsegment  E O
drawsegment  A P
drawsegment  A E
drawsegment  A B
drawcircle  k
drawdashline  p
drawsegment  E A
drawsegment  E B



Nome Utilizador: 
point  E 35 20
point  O 35 45
circle  k O E
oncircle  A  O E
oncircle  B  O E
line  ab A B
line  oe O E
perp  ep E oe
intersec  P ep ab
line  ae A E
line  be B E
bis  p E P A
intersec  C ae p
intersec  D be p
prove { equal { segment E D } { segment E C } }


Nome do Teorema: chou88


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Assume the incircle of $\triangle ABC$ touches the 
 sides $BC$, $AC$ and $AB$ in points 
 $D$, $E$ and $F$ respectively. 
 Show that lines $AD$, $BE$ and $CF$ meet at the Gergonne 
 point $G$ of the triangle. 
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  100 100
point  A 20 20
point  B 80 20
point  C 70 70
bis  ba B A C
bis  bb C B A
bis  bc A C B
line  a B C
line  b A C
line  c A B
intersec  I ba bb
foot  D I a
foot  E I b
foot  F I c
circle  k I F
line  ad A D
line  be B E
line  cf C F
intersec  G ad be
cmark_b  A
cmark_b  B
cmark_t  C
cmark_b  F
cmark_lt  E
cmark_rt  D
cmark_lt  G
drawcircle  k
drawsegment  A B
drawsegment  C B
drawsegment  C A
drawdashsegment  A D
drawdashsegment  B E
drawdashsegment  C F



Referência BiBTeX: 
point  A 20 20
point  B 80 20
point  C 70 70
bis  ba B A C
bis  bb C B A
bis  bc A C B
line  a B C
line  b A C
line  c A B
intersec  I ba bb
foot  D I a
foot  E I b
foot  F I c
line  ad A D
line  be B E
line  cf C F
intersec  G ad be
prove { collinear C F G }


Nome Utilizador: 


Nome do Teorema: gpreda1


Teorema: 
 Adams  Circle



Descrição Breve: 


Observações: 


Código GCLC: 
dim  130 130
point  A 20 20
point  B 110 20
point  C 95 105
bis  ba B A C
bis  bb C B A
bis  bc A C B
line  a B C
line  b A C
line  c A B
intersec  I ba bb
foot  D I a
foot  E I b
foot  F I c
circle  k I F
line  ad A D
line  be B E
line  cf C F
intersec  G ad be
line  de D E
line  df D F
line  ef E F
parallel  gpq G de
parallel  grs G df
parallel  gtu G ef
intersec  P gpq b
intersec  Q gpq a
intersec  R grs a
intersec  S grs c
intersec  T gtu c
intersec  U gtu b
cmark_l  I
cmark_b  A
cmark_b  B
cmark_t  C
cmark_b  F
cmark_lt  E
cmark_rt  D
cmark_rb  G
cmark_lt  P
cmark_r  Q
cmark_rt  R
cmark_b  S
cmark_b  T
cmark_lt  U
drawcircle  k
drawsegment  A B
drawsegment  C B
drawsegment  C A
drawsegment  A D
drawsegment  B E
drawsegment  C F
drawdashsegment  P Q
drawdashsegment  R S
drawdashsegment  T U
drawsegment  D E
drawsegment  D F
drawsegment  F E
circle  k1 I S
drawdashcircle  k1



Código GCLCprover: 
point  A 20 20
point  B 110 20
point  C 95 105
bis  ba B A C
bis  bb C B A
bis  bc A C B
line  a B C
line  b A C
line  c A B
intersec  I ba bb
foot  D I a
foot  E I b
foot  F I c
line  ad A D
line  be B E
line  cf C F
intersec  G ad be
line  de D E
line  df D F
line  ef E F
parallel  gpq G de
parallel  grs G df
parallel  gtu G ef
intersec  Q gpq a
intersec  S grs c
intersec  T gtu c
prove { equal { segment I S } { segment I T } }
prove { equal { segment I S } { segment I Q } }


Referência BiBTeX: 


Nome Utilizador: gpreda1


Nome do Teorema: 
 Example 5.1 


Teorema: \begin{geothm}
 Let $ABCD$ be a square. $CG$ is parallel
 to the diagonal $BD$. Point $E$ is on
 $CG$ such that $BE \cong BD$. $F$ is the
 intersection of $BE$ and $DC$. Show that
 $DF \cong DE$.
\end{geothm}

Descrição Breve: 


Observações: 
dim  110 60
point  A 59 20
point  B 78 20
line  ab A B
perp  p A ab
circle  k A B
intersec2  D D_1 k p
perp  q D p
perp  r B ab
intersec  C q r
line  d B D
parallel  c C d
circle  k1 B D
intersec2  E E_1 k1 c
line  be B E
line  cd C D
intersec  F be cd
line  be1 B E_1
intersec  F_1 be1 cd
cmark_b  A
cmark_b  B
cmark_t  D
cmark_t  C
cmark_t  E
cmark_lb  F
cmark_rt  E_1
cmark_t  F_1
drawsegment  A B
drawsegment  A D
drawsegment  C B
drawsegment  F_1 C
drawsegment  B D
drawsegment  B E
drawsegment  E D
drawsegment  C E
drawsegment  E E_1
drawsegment  F_1 E_1
drawdashsegment  D E_1



Código GCLC: 
point  A 59 20
point  B 78 20
line  ab A B
perp  p A ab
circle  k A B
intersec2  D D_1 k p
perp  q D p
perp  r B ab
intersec  C q r
line  d B D
parallel  c C d
circle  k1 B D
intersec2  E E_1 k1 c
line  be B E
line  cd C D
intersec  F be cd
prove { equal { segment D F } { segment D E } }


Código GCLCprover: chou88


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Example 95 


Nome do Teorema: \begin{geothm}
 Let $M$ be the midpoint of the arc $AB$
 of circle $(O)$, $D$ be the midpoint of
 $AB$. he perpendicular through $M$ is drawn
 to the tangent of the circle at $A$
 meeting that tangent at $E$. Show $ME \cong MD$.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  100 100
point  D 50 33
point  O 50 55
line  d O D
point  M 50 90
circle  k O M
perp  a D d
intersec2  B A k a
line  p A O
perp  q A p
parallel  r M p
intersec  E r q
cmark_t  E
cmark_lb  A
cmark_rb  B
cmark_r  O
cmark_b  D
cmark_t  M
drawcircle  k
drawsegment  A B
drawsegment  A O
drawsegment  M D
drawsegment  E M
drawsegment  A E



Observações: 
point  D 50 33
point  O 50 55
line  d O D
point  M 50 90
circle  k O M
perp  a D d
intersec2  B A k a
line  p A O
perp  q A p
parallel  r M p
intersec  E r q
prove { equal { segment M E } { segment M D } }


Código GCLC: chou88


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Example 97 


Nome Utilizador: \begin{geothm}
 Let $G$ be a point on the circle $(O)$ with
 diameter $BC$, $A$ be the midpoint of the arc
 $BG$. $AD \perp BC$. $E = AD \cap BG$ and
 $F = AC \cap BG$. Show that
 $AE \cong BE \cong EF$.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  B 10 50
point  C 90 50
midpoint  O B C
circle  k O B
oncircle  G  O B
med  m B G
intersec2  A A_1 k m
line  b B C
perp  d A b
line  g B G
line  c A C
intersec  E d g
intersec  D d b
intersec  F g c 
prove { equal { segment A E } { segment B E } }
prove { equal { sratio B E E F } { 1 } }


Observações: chou88


Código GCLC: gpreda1


Código GCLCprover: 
 Example 102 


Referência BiBTeX: \begin{geothm}
 The two tangents to the circumcircle
 of $ABC$ at $A$ and $C$ meet at $E$.
 The mediator of $BC$ meet $AB$
 at $D$. Show that $DE \parallel BC$.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  110 85
point  A 35 25
point  B 90 25
point  C 58 68
med  ma B C
med  mc A B
intersec  O ma mc
circle  k O A
line  a B C
intersec  H ma a
line  c A B
intersec  D ma c
line  ao A O
line  co C O
perp  pa A ao
perp  pc C co
intersec  E pa pc
cmark_t  E
cmark_b  D
cmark_rb  O
drawcircle  k
cmark_b  A
cmark_b  B
cmark_t  C
cmark_rt  H
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  A O
drawsegment  C O
drawsegment  A E
drawsegment  C E
drawsegment  D H
drawdashsegment  D E



Descrição Breve: chou88


Observações: gpreda1


Código GCLC: 
 Example 140 


Código GCLCprover: \begin{geothm}
 The product of two sides of a triangle
 is equal to the altitude to the third side
 multiplied by the circumdiameter.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
circle  k O A
foot  F C c
cmark_b  F
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_t  O
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C F
drawsegment  A O
drawcircle  k



Teorema: 
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
circle  k O A
foot  F C c
prove { equal { mult { segment A C } { segment B C } } { mult { 4 } { mult { segment O A } { segment C F } } } }


Descrição Breve: gpreda1


Observações: 
 Example 138 


Código GCLC: \begin{geothm}
 The internal and external bisectors of an
 angle of a triangle pass through the ends
 of the circumdiameters which is a perpendicular
 to the side opposite the vertex considered.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
circle  k O A
perp  l O c
intersec2  L1 L l k
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_t  O
cmark_b  L
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  O L
drawsegment  C L
drawcircle  k



Nome do Teorema: 
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
circle  k O A
perp  l O c
intersec2  L1 L l k
prove { equal { angle A C L } { angle L C B } }


Teorema: chou88


Descrição Breve: 
 Example 101 


Observações: \begin{geothm}
 On the hypotenuse $AB$ of right triangle
 $ABC$ a square $ABFE$ is erected.
 Let $P$ be the intersection of the diagonals
 $AF$ and $BE$ of $ABFE$. Show that:
 $\angle A C P = \angle P C B$.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  110 110
point  C 40 30
point  B 65 30
point  A 40 65
line  c A B
perp  f B c
perp  e A c
circle  k A B
intersec2  E E_1 k e
perp  ef E e
intersec  F ef f
line  af A F
line  be B E
intersec  P af be
perp  ef1 E_1 e
intersec  F_1 ef1 f
line  af1 A F_1
line  be1 B E_1
intersec  P_1 af1 be1
cmark_lb  C
cmark_b  B
cmark_lt  A
cmark_t  E
cmark_rb  F
cmark_rb  P
cmark_t  E_1
cmark_rb  F_1
cmark_lb  P_1
drawsegment  A B
drawsegment  C B
drawsegment  A C
drawsegment  A E
drawsegment  A F
drawsegment  B F
drawsegment  E F
drawsegment  B E
drawsegment  C P
drawdashsegment  A E_1
drawdashsegment  A F_1
drawdashsegment  B F_1
drawdashsegment  E_1 F_1
drawdashsegment  B E_1
drawdashsegment  C P_1



Nome Utilizador: 
point  C 40 30
point  B 65 30
point  A 40 65
line  c A B
perp  f B c
perp  e A c
circle  k A B
intersec2  E E_1 k e
perp  ef E e
intersec  F ef f
line  af A F
line  be B E
intersec  P af be
prove { equal { angle A C P } { angle P C B } }


Nome do Teorema: chou88


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 A line parallel to the base of trapezoid
 $ABCD$ meet ifs two sides and two diagonals
 at $H$, $G$, $F$ and $E$. Show that:
 $EF \cong GH$.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  110 110
point  A 20 20
point  B 90 20
point  D 45 90
point  C 85 90
point  H 31.95 53.46
line  a A B
parallel  h H a
line  b B C
intersec  E h b
line  d B D
line  c C A
intersec  G c h
intersec  F d h
cmark_b  A
cmark_b  B
cmark_t  C
cmark_t  D
cmark_l  H
cmark_r  E
cmark_rb  G
cmark_lb  F
drawsegment  A B
drawsegment  C B
drawsegment  C D
drawsegment  A D
drawsegment  A C
drawsegment  B D
drawsegment  H E



Referência BiBTeX: 
point  A 20 20
point  B 90 20
point  D 45 90
point  C 85 90
online  H  A D
line  a A B
parallel  h H a
line  b B C
intersec  E h b
line  d B D
line  c C A
intersec  G c h
intersec  F d h
prove { equal { sratio H G F E } { 1 } }


Nome Utilizador: chou88


Nome do Teorema: gpreda1


Teorema: 
 Example 110 


Descrição Breve: 


Observações: 


Código GCLC: 
dim  120 110
point  A 20 40
point  B 85 40
point  C 70 90
point  D 40.26 60.26
line  b A C
parallel  b1 B b
line  c A B
parallel  c1 D c
intersec  D1 b1 c1
line  a B C
circle  k B D1
intersec2  E1 E k a
line  de D E
intersec  F de c
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_lt  D
cmark_b  E
cmark_lb  F
drawsegment  A B
drawsegment  C E
drawsegment  A C
drawdashsegment  B D1
drawdashsegment  D D1
drawsegment  D E



Código GCLCprover: 
point  A 20 40
point  B 85 40
point  C 70 90
online  D  A C
line  b A C
parallel  b1 B b
line  c A B
parallel  c1 D c
intersec  D1 b1 c1
line  a B C
circle  k B D1
intersec2  E1 E k a
line  de D E
intersec  F de c
prove { equal { mult { segment F D } { segment A C } } { mult { segment E F } { segment B C } } }


Referência BiBTeX: chou88


Nome Utilizador: gpreda1


Nome do Teorema: 
 Example 121 


Teorema: \begin{geothm}
 Two parallel line $AE$, $BD$ through the vertices
 $A$, $B$ of the triangle $ABC$ meet a line through
 the vertex $C$ in the points $E$, $D$. If the
 parallel through $E$ to $BC$ meets $AB$ in $F$,
 show that $DF$ is parallel to $AC$.
\end{geothm}

Descrição Breve: 


Observações: 
dim  110 110
point  A 20 35
point  B 90 35
point  C 70 90
point  D 50 20
line  bd B D
parallel  ae A bd
line  cd C D
intersec  E ae cd
line  bc B C
parallel  ef E bc
line  ab A B
intersec  F ab ef
cmark_b  A
cmark_b  B
cmark_t  C
cmark_b  D
cmark_r  E
cmark_rb  F
drawsegment  A B
drawsegment  C D
drawsegment  C B
drawsegment  A E
drawsegment  E F
drawdashsegment  A C
drawdashsegment  D F



Código GCLC: 
point  A 20 35
point  B 90 35
point  C 70 90
point  D 50 20
line  bd B D
parallel  ae A bd
line  cd C D
intersec  E ae cd
line  bc B C
parallel  ef E bc
line  ab A B
intersec  F ab ef
prove { parallel A C D F }


Código GCLCprover: chou88


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Example 108 


Nome do Teorema: \begin{geothm}
 Let $C$ be the midpoint of the arc $AB$
 of circle $(O)$. $D$ is a point on the circle.
 $E = AB \cap CD$. Show that:
 $CA^2 = CE \cdot CD$.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  110 110
point  A 20 30
point  B 90 30
point  X 25 34
line  ab A B
line  ax A X
med  c A B
intersec  O c ax
intersec  M c ab
circle  k O A
intersec2  C C1 k c
point  D 39.67 15.88
line  cd C D
intersec  E ab cd
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_r  O
cmark_b  M
cmark_lb  D
cmark_lt  E
drawcircle  k
drawsegment  A B
drawsegment  C B
drawsegment  A C
drawsegment  C M
drawsegment  C D



Observações: 
point  A 20 30
point  B 90 30
point  X 25 34
line  ab A B
line  ax A X
med  c A B
intersec  O c ax
intersec  M c ab
circle  k O A
intersec2  C C1 k c
oncircle  D  O A
line  cd C D
intersec  E ab cd
prove { equal { mult { segment C A } { segment C A } } { mult { segment C E } { segment C D } } }


Código GCLC: chou88


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Example 114 


Nome Utilizador: \begin{geothm}
 Through point $F$ on the circle with
 diameter $AB$ a tangent to the circle
 is drawn meeting the two lines,
 perpendicular to $AB$ at $A$ and $B$,
 at $D$ and $E$. Show that:
 $OA^2 = DF \cdot EF$.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  A 20 45
point  B 90 45
midpoint  O A B
circle  k O A
oncircle  F  O A
line  ab A B
line  f F O
perp  de F f
perp  ad A ab
perp  be B ab
intersec  D ad de
intersec  E be de
prove { equal { mult { segment O A } { segment O A } } { mult { segment D F } { segment E F } } }


Observações: chou88


Código GCLC: gpreda1


Código GCLCprover: 
 Example 117 


Referência BiBTeX: \begin{geothm}
 From a point $A$ two lines are drawn tangent to
 circle $(O)$ at $B$ and $C$. From a point $P$ on
 the circle peprendiculars are drawn to $BC$, $AB$
 and $AC$. Let $D$, $F$, $E$ be the feet. Show that:
 $PD^2 = PE \cdot PF$.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  110 90
point  A 20 20
point  B 70 20
line  ab A B
point  X 60 70
line  ac A X
bis  s B A X
perp  b B ab
intersec  O b s
foot  C O ac
circle  k O B
point  P 46.05 42.01
line  bc B C
foot  F P ab
foot  E P ac
foot  D P bc
cmark_b  A
cmark_b  B
cmark_rt  O
cmark_lt  C
cmark_rt  D
cmark_b  F
cmark_lt  E
cmark_lb  P
drawcircle  k
drawsegment  A B
drawsegment  A C
drawsegment  O B
drawsegment  O C
drawsegment  C B
drawsegment  P F
drawsegment  P E
drawsegment  P D



Descrição Breve: chou88


Observações: gpreda1


Código GCLC: 
 Example 127 


Código GCLCprover: \begin{geothm}
 Of $Q$, $R$ are the projections of a point $M$
 of the internal bisector $AM$ of the angle $A$
 of the triangle $ABC$ upon the sides $AC$, $AB$,
 show that the perpendicular $MP$ from $M$ upon
 $BC$ meets $QR$ in the point $N$ on the median
 $AA_1$ of $ABC$.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  150 110
point  A 65 20
point  B 120 20
point  C 50 90
line  a C B
bis  s B A C
intersec  X s a
point  M 102.91 66.89
line  b A C
line  c A B
foot  Q M b
foot  R M c
midpoint  A_1 B C
line  a1 A A_1
line  qr Q R
intersec  N qr a1
point  B1 10 20
bis  s1 B1 A C
point  M_1 48.20 33.60
foot  Q_1 M_1 b
foot  R_1 M_1 c
line  qr1 Q_1 R_1
intersec  N_1 qr1 a1
cmark_b  A
cmark_b  B
cmark_l  C
cmark_t  M
cmark_lb  Q
cmark_b  R
cmark_rt  A_1
cmark_t  N
cmark_lt  M_1
cmark_r  Q_1
cmark_b  R_1
cmark_t  N_1
drawsegment  R_1 B
drawsegment  C B
drawsegment  A C
drawsegment  A A_1
drawsegment  M N
drawsegment  M Q
drawsegment  A M
drawsegment  M R
drawdashsegment  Q R
drawdashsegment  M_1 N_1
drawdashsegment  M_1 Q_1
drawdashsegment  A M_1
drawdashsegment  M_1 R_1
drawdashsegment  N_1 R_1



Teorema: 
point  A 65 20
point  B 120 20
point  C 50 90
line  a C B
bis  s B A C
intersec  X s a
online  M  A X
line  b A C
line  c A B
foot  Q M b
foot  R M c
midpoint  A_1 B C
line  a1 A A_1
line  qr Q R
intersec  N qr a1
prove { perpendicular B C M N }


Descrição Breve: gpreda1


Observações: 
 Example 133 


Código GCLC: \begin{geothm}
 If $M$, $N$ are points on the sides $AC$,
 $BC$ of a triangle $ABC$ and the lines
 $AN$, $BM$ intersect on the altitude $CD$,
 show that $CD$ is the bisector of the 
 angle $\angle MDN$.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  110 110
point  A 20 20
point  B 90 20
point  C 70 80
line  c A B
line  a B C
line  b A C
foot  D C c
point  J 70 50
line  aj A J
line  bj B J
intersec  M bj b
intersec  N aj a
cmark_b  A
cmark_b  B
cmark_t  C
cmark_lt  M
cmark_rt  N
cmark_b  D
cmark_lb  J
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C D
drawsegment  M B
drawsegment  A N
drawsegment  D M
drawsegment  D N



Nome do Teorema: 
point  A 20 20
point  B 90 20
point  C 70 80
line  c A B
line  a B C
line  b A C
foot  D C c
online  J  C D
line  aj A J
line  bj B J
intersec  M bj b
intersec  N aj a
prove { equal { angle M D C } { angle C D N } }


Teorema: chou88


Descrição Breve: 
 Example 136 


Observações: \begin{geothm}
 Show that the internal (or external) bisector of an
 angle of a triangle is divided harmonically by the
 feet of the perpendiculars dropped upon it from
 the two other vertices of the triangle.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 70 80
line  c A B
bis  s A C B
foot  K B s
foot  J A s
intersec  F c s
towards  A_1 A C 2
bis  s_1 A_1 C B
foot  K_1 B s_1
foot  J_1 A s_1
intersec  F_1 c s_1
cmark_b  A
cmark_b  B
cmark_t  C
cmark_rb  F
cmark_rb  J
cmark_l  K
cmark_rb  F_1
cmark_t  J_1
cmark_t  K_1
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C J
drawsegment  B K
drawsegment  A J
drawdashsegment  F_1 J_1
drawdashsegment  B K_1
drawdashsegment  A J_1
drawdashsegment  C A_1



Nome Utilizador: 
point  A 20 30
point  B 90 30
point  C 70 80
line  c A B
bis  s A C B
foot  K B s
foot  J A s
intersec  F c s
prove { harmonic C F J K }


Nome do Teorema: chou88


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 The angle between the circumdiameter and the
 altitude issued from the same vertex of a
 triangle is bisected by the bisector of angle
 of the triangle at the same vertex considered.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
bis  s A C B
intersec  F s c
foot  D C c
circle  k O A
towards  A_1 A C 2
bis  s1 A_1 C B
drawdashline  s1
towards  O_1 O C 2
drawdashsegment  C O_1
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_b  F
cmark_b  D
cmark_b  O
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C D
drawsegment  C F
drawsegment  C O
drawcircle  k



Referência BiBTeX: 
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
bis  s A C B
intersec  F s c
foot  D C c
circle  k O A
prove { equal { angle O C F } { angle F C D } }


Nome Utilizador: chou88


Nome do Teorema: gpreda1


Teorema: 
 Example 141 


Descrição Breve: 


Observações: 


Código GCLC: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
circle  k O A
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_t  O
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  A O
drawcircle  k



Código GCLCprover: 
point  A 20 30
point  B 90 30
point  C 73 85
line  c A B
med  mc B C
med  mb A C
intersec  O mc mb
circle  k O A
prove { equal { mult { 16 } { mult { signed_area3 A B C } { mult { signed_area3 A B C } { segment A O } } } } { mult { segment A B } { mult { segment A C } { segment B C } } } } }



Referência BiBTeX: chou88


Nome Utilizador: gpreda1


Nome do Teorema: 
 Example 132 


Teorema: \begin{geothm}
 Let $M$ be the midpoint of chord $AB$ of a circle
 with center $O$. On $OM$ as diameter draw another
 circle, and at any point $T$ of this circle
 draw a tangent to it meeting the outer circle in $E$.
 Prove that:
 $AE^2 + BE^2 = 4ET^2$.
\end{geothm}

Descrição Breve: 


Observações: 
dim  110 110
point  A 30 30
point  B 80 30
med  m A B
point  X 35 34
line  ax A X
intersec  O ax m
line  ab A B
intersec  M m ab
circle  k O A
midpoint  N O M
circle  k1 N M
point  T 45.46 37.01
line  tn T N
perp  t T tn
intersec2  E E1 t k
cmark_lb  A
cmark_rb  B
cmark_lt  O
cmark_b  M
cmark_r  N
cmark_lb  T
cmark_lt  E
drawsegment  A B
drawsegment  A E
drawsegment  E B
drawsegment  O M
drawsegment  T E
drawsegment  T N
drawcircle  k
drawcircle  k1



Código GCLC: 
point  A 30 30
point  B 80 30
med  m A B
point  X 35 34
line  ax A X
intersec  O ax m
line  ab A B
intersec  M m ab
circle  k O A
midpoint  N O M
circle  k1 N M
oncircle  T  N M
line  tn T N
perp  t T tn
intersec2  E E1 t k
prove { equal { mult { 4 } { segment E T } } { sum { segment A E } { segment B E } } }


Código GCLCprover: chou88


Referência BiBTeX: gpreda1


Nome Utilizador: 
 General Menelaus s Theorem



Nome do Teorema: \begin{geothm}
 Line $p$ intersects lines $AB$, $BC$, $CD$ and $DA$
 at points $A_1$, $B_1$, $C_1$ and $D_1$ respectively.
 Show that:
 $\frac{\overline{AA_1}}{\overline{A_1B}} \cdot
 \frac{\overline{BB_1}}{\overline{B_1C}} \cdot
 \frac{\overline{CC_1}}{\overline{C_1D}} \cdot
 \frac{\overline{DD_1}}{\overline{D_1A}} \cdot
 = 1$
\end{geothm}

Teorema: 


Descrição Breve: 
dim  100 100
point  A 20 20
point  B 80 20
point  C 70 60
point  D 37 74
line  ab A B
line  bc B C
line  da D A
line  cd C D
point  A_1 85 20
point  B_1 77.5 32.5
line  a1b1 A_1 B_1
intersec  C_1 a1b1 cd
intersec  D_1 a1b1 da
cmark_rb  A
cmark_b  B
cmark_r  C
cmark_l  D
cmark_rt  A_1
cmark_r  B_1
cmark_rt  C_1
cmark_r  D_1
drawsegment  A B
drawsegment  C B
drawsegment  C D
drawsegment  A D
drawdashline  a1b1
drawline  da
drawline  ab



Observações: 
point  A 20 20
point  B 80 20
point  C 70 60
point  D 37 74
line  ab A B
line  bc B C
line  da D A
line  cd C D
online  A_1  A B
online  B_1  B C
line  a1b1  A_1  B_1 
intersec  C_1  a1b1  cd
intersec  D_1  a1b1  da
prove { equal { mult { mult { mult { sratio A A_1 A_1 B } { sratio B B_1 B_1 C } } { sratio C C_1 C_1 D } } { sratio D D_1 D_1 A } } 1 }


Código GCLC: 


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Pappus  Theorem



Nome Utilizador: \begin{geothm}
 Let $ABC$ and $A_1B_1C_1$ be two lines such that
 $AB_1 \parallel BA_1$ and $BC_1 \parallel CB_1$.
 Show that $AC_1 \parallel CA_1$.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  A 20 10
point  B 90 10
online  C  A B 
point  A_1 55 50
point  A_2 45 43
line  A_1A_2 A_1 A_2
line  A_1B A_1 B
parallel  AB_1 A A_1B
intersec  B_1 A_1A_2 AB_1
line  CB_1  C  B_1
parallel  BC_1  B CB_1 
intersec  C_1  A_1A_2 BC_1 
prove { parallel A_1 C A C_1 }



Observações: 


Código GCLC: gpreda1


Código GCLCprover: 
 Pappus  Theorem (second version)



Referência BiBTeX: \begin{geothm}
 Let $ABC$ and $A_1B_1C_1$ be two lines and
 $P = AB_1 \cap A_1B$, $Q = AC_1 \cap A_1C$,
 $S = BC_1 \cap B_1C$. Then $P$, $Q$ and $S$
 are collinear.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  100 100
point  A 20 20
point  B 50 20
point  C 77 20
point  A_1 25 50
point  B_1 52 60
point  C_1 65.50 65.00
line  ab1 A B_1
line  ba1 B A_1
intersec  P ab1 ba1
line  ac1 A C_1
line  ca1 C A_1
intersec  Q ac1 ca1
line  bc1 B C_1
line  cb1 C B_1
intersec  S bc1 cb1
line  p A B
line  q A_1 B_1
line  r P Q   
cmark_b  A
cmark_b  B
cmark_b  C
cmark_t  A_1
cmark_t  B_1
cmark_t  C_1
cmark_b  P
cmark_b  Q
cmark_r  S
drawline  p
drawline  q
drawdashline  r
drawsegment  A B_1
drawsegment  A_1 B
drawsegment  A C_1
drawsegment  A_1 C
drawsegment  C B_1
drawsegment  C_1 B



Descrição Breve: 


Observações: gpreda1


Código GCLC: 
 Example 145 


Código GCLCprover: \begin{geothm}
 Show that the distance of a point on a
 median of triangle from the sides including
 the median are inversely proportional to these
 sides, ie:
 $NK : NJ = BC : AC$.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 40 88
midpoint  F A B
line  a B C
line  b A C
point  N 49.18 52.51
foot  K N b
foot  J N a
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_b  F
cmark_lt  K
cmark_rt  J
cmark_lb  N
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  C F
drawsegment  N K
drawsegment  N J



Teorema: 
point  A 20 30
point  B 90 30
point  C 40 88
midpoint  F A B
line  a B C
line  b A C
online  N  C F
foot  K N b
foot  J N a
prove { equal { mult { segment N K } { segment A C } } { mult { segment N J } { segment B C } } }



Descrição Breve: gpreda1


Observações: 
 Example 150 


Código GCLC: \begin{geothm}
 If $G$ is a centroid of triangle $ABC$,
 show that for any given point $M$:
 $3 \cdot MG^2 + AG^2 + BG^2 + CG^2 = AM^2 + BM^2 + CM^2$.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 40 88
point  M 93 90
midpoint  F A B
midpoint  E A C
midpoint  D B C
line  cf C F
line  ad A D
intersec  G ad cf
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_t  M
cmark_lt  G
cmark_b  F
cmark_lt  E
cmark_rt  D
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawsegment  M B
drawsegment  C M
drawsegment  C F
drawsegment  A D
drawsegment  B E
drawsegment  G M



Nome do Teorema: 
point  A 20 30
point  B 90 30
point  C 40 88
point  M 93 90
midpoint  F A B
midpoint  E A C
midpoint  D B C
line  cf C F
line  ad A D
intersec  G ad cf
prove { equal { sum { mult { segment M G } { 3 } } { sum { segment A G } { sum { segment C G } { segment B G } } } } { sum { segment A M } { sum { segment B M } { segment C M } } } }



Teorema: chou88


Descrição Breve: 
 Example 155 


Observações: \begin{geothm}
 Two tritangent centers divide the bisector
 on which they are located, harmonically
 (ie, points $C$, $D$, $I$ and $I_c$
 are harmonic).
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  120 110
point  A 20 63
point  B 85 63
point  C 28 98
bis  sc A C B
bis  sa B A C
perp  sa_1 A sa
line  c A B
intersec  I sa sc
intersec  I_c sa_1 sc
intersec  D sc c
foot  X I c
foot  Y I_c c
circle  k I X
circle  k1 I_c Y
towards  A1 C A 10
towards  B1 C B 10
drawdashsegment  A A1
drawdashsegment  B B1
drawdashcircle  k
drawdashcircle  k1
cmark_lt  A
cmark_rt  B
cmark_t  C
cmark_rt  I
cmark_lb  D
cmark_rt  I_c
drawsegment  A C
drawsegment  C B
drawsegment  A B
drawsegment  C I_c
drawsegment  A I
drawsegment  B I
drawsegment  A I_c



Nome Utilizador: 
point  A 20 63
point  B 85 63
point  C 28 98
bis  sc A C B
bis  sa B A C
perp  sa_1 A sa
line  c A B
intersec  I sa sc
intersec  I_c sa_1 sc
intersec  D sc c
prove { harmonic C D I I_c }



Nome do Teorema: chou88


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Show that in any given triangle the ratio
 between one triangle side and a diameter of
 circumscribed circle is equal to the sine
 of opposite angle, ie:
 $\frac{a}{2R} = \sin \alpha$
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 70 90
med  mc A B
med  mb A C
intersec  O mc mb
circle  k O A
line  c A B
foot  D C c
cmark_b  A
cmark_b  B
cmark_t  C
cmark_b  O
drawcircle  k
cmark_b  D
drawsegment  O C
drawsegment  C D
drawsegment  A B
drawsegment  C B
drawsegment  A C
point  a 83.00 58.40
printat  a { a }
point  b 42.40 61.00
printat  b { b }
point  c 57.90 28.20
printat  c { c }
point  al 30.00 34.10
printat  al { \alpha }
point  r 60.00 70.90
printat  r { R }
point  x 40 30
drawarc  A x 50
point  xx 41 30
drawarc  A xx 50



Referência BiBTeX: 
point  A 20 30
point  B 90 30
point  C 70 90
med  mc A B
med  mb A C
intersec  O mc mb
circle  k O A
line  c A B
foot  D C c
prove { equal { mult { 4 } { mult { segment A O } { segment C D } } } { mult { segment B C } { segment C A } } }


Nome Utilizador: 


Nome do Teorema: gpreda1


Teorema: 
 The Law of Cosine



Descrição Breve: 


Observações: 


Código GCLC: 
dim  110 110
point  A 20 30
point  B 90 30
point  C 70 90
line  c A B
foot  D C c
cmark_b  A
cmark_b  B
cmark_b  D
cmark_t  C
drawsegment  C D
drawsegment  A B
drawsegment  C B
drawsegment  A C
point  a 83.00 58.40
printat  a { a }
point  b 42.40 61.00
printat  b { b }
point  c 57.90 28.20
printat  c { c }
point  al 30.00 34.10
printat  al { \alpha }
point  x 40 30
drawarc  A x 50
point  xx 41 30
drawarc  A xx 50



Código GCLCprover: 
point  A 20 30
point  B 90 30
point  C 70 90
line  c A B
foot  D C c
prove { equal { mult { mult { sum { mult { -1 } { segment C B } } { sum { segment A C } { segment B A } } } { sum { mult { -1 } { segment C B } } { sum { segment A C } { segment B A } } } } { segment A C } } { mult { 4 } { mult { segment A C } { mult { segment B A } { segment A D } } } } }


Referência BiBTeX: 


Nome Utilizador: gpreda1


Nome do Teorema: 
 Euler s Formula



Teorema: \begin{geothm}
 If $R$ and $r$ are half diameters of circumscribed
 and inscribed circles of a trougle and $d$ is a
 distance between them, show that:
 $d^2 = R(R - 2r)$.
\end{geothm}

Descrição Breve: 


Observações: 
dim  105 90
point  A 20 33
point  B 85 33
point  C 68 75
bis  sc A C B
bis  sa B A C
intersec  I sa sc
med  mc A B
med  mb A C
intersec  O mc mb
circle  k O A
line  c A B
foot  D I c
circle  k1 I D
drawcircle  k1
drawcircle  k
cmark_b  O
cmark_lb  A
cmark_rb  B
cmark_t  C
cmark_rt  I
cmark_b  D
drawsegment  A O
drawsegment  I D
drawsegment  A C
drawsegment  C B
drawsegment  A B
drawdashsegment  I O
point  x 56.00 48.00
printat  x { d }
point  y 38.90 41.50
printat  y { R }
point  z 63.30 41.30
printat  z { r }



Código GCLC: 
point  A 20 33
point  B 85 33
point  C 68 75
bis  sc A C B
bis  sa B A C
intersec  I sa sc
med  mc A B
med  mb A C
intersec  O mc mb
circle  k O A
line  c A B
foot  D I c
circle  k1 I D
prove { equal { mult { sum { mult { -1 } { segment I O } } { segment A O } } { sum { mult { -1 } { segment I O } } { segment A O } } } { mult { 4 } { mult { segment A O } { segment I D } } } }


Código GCLCprover: 


Referência BiBTeX: gpreda1


Nome Utilizador: 
 Example 172 


Nome do Teorema: \begin{geothm}
 Show that the area of a right triangle is equal
 to the product of the two segments into
 which the hypothenuse is divided by its point
 of contact with incircle.
\end{geothm}

Teorema: 


Descrição Breve: 
dim  90 90
point  A 20 20
point  B 70 20
line  c A B
point  X 30 65
line  x B X
perp  b A c
intersec  C x b
bis  sa B A C
bis  sb A B C
intersec  I sa sb
line  a B C
foot  D I a
circle  k I D
drawcircle  k
cmark_rt  D
cmark_lt  I
cmark_b  A
cmark_b  B
cmark_t  C
drawsegment  A C
drawsegment  C B
drawsegment  A B
foot  E I c
foot  F I b
drawdashsegment  I E
drawdashsegment  I F
drawdashsegment  I D



Observações: 
point  A 20 20
point  B 70 20
line  c A B
point  X 30 65
line  x B X
perp  b A c
intersec  C x b
bis  sa B A C
bis  sb A B C
intersec  I sa sb
line  a B C
foot  D I a
circle  k I D
prove { equal { mult { segment C D } { segment B D } } { mult { signed_area3 A B C } { signed_area3 A B C } } }



Código GCLC: chou88


Código GCLCprover: gpreda1


Referência BiBTeX: 
 Example 174 


Nome Utilizador: \begin{geothm}
 The segment of the altitude extended
 between the orthocenter and the second
 point of intersection with the circumcircle
 is bisected by the corresponding side of the triangle.
\end{geothm}

Nome do Teorema: 


Teorema: 


Descrição Breve: 
point  A 20 30
point  K 90 30
point  D 70 30
line  a A K
perp  c D a
med  o A K
point  X 24 33
line  x A X
intersec  O o x
circle  k O A
intersec2  C B c k
line  b A B
foot  E C b
line  hc C E
intersec  H hc a
prove { equal { sratio H D D K } { 1 } }



Observações: chou88


Código GCLC: gpreda1


Código GCLCprover: 
 Example 219 


Referência BiBTeX: \begin{geothm}
 Show that the mediator of the bisector $CU$ of the
 triangle $ABC$, the perpendicular to $AB$ at $U$,
 and the circumdiameter of $ABC$ passing through $C$
 are concurrent.
\end{geothm}

Nome Utilizador: 


Nome do Teorema: 


Teorema: 
dim  110 100
point  A 20 30
point  B 90 30
point  C 75 85
med  mc A B
med  mb A C
intersec  O mc mb
circle  k O A
bis  s_c A C B
line  c A B
intersec  U s_c c
line  p C O
med  m C U
intersec  P p m
drawdashline  m
drawcircle  k
cmark_lt  P
cmark_b  U
cmark_l  O
cmark_lb  A
cmark_rb  B
cmark_t  C
drawdashsegment  P U
drawsegment  C O
drawsegment  C U
drawsegment  A B
drawsegment  A C
drawsegment  C B



Descrição Breve: chou88


Observações: gpreda1


Código GCLC: 
 Triangle median line



Código GCLCprover: \begin{geothm}
 Let $A_1$ and $B_1$ be midpoints of segments
 $AC$ and $BC$ in triangle $ABC$.
 Show that $A_1B_1$ is parallel to $AB$
 and half of that segment.
\end{geothm}

Referência BiBTeX: 


Nome Utilizador: 


Nome do Teorema: 
dim  80 50
point  A 20 10
point  B 70 10
point  C 35 40
drawsegment  A B
drawsegment  A C
drawsegment  B C
midpoint  B_1 B C
midpoint  A_1 A C
drawsegment  A_1 B_1
midpoint  M A B
cmark_b  M
drawdashsegment  A_1 M
drawdashsegment  B_1 M
cmark_b  A
cmark_b  B
cmark_t  C
cmark_lt  A_1
cmark_rt  B_1



Teorema: 
point  A 20 10
point  B 70 10
point  C 35 40
midpoint  B_1 B C
midpoint  A_1 A C
prove { parallel A_1 B_1 A B }



Descrição Breve: gpreda1


Observações: 
 Projection Keeps Cross Ratio



Código GCLC: \begin{geothm}
 Cross ratio of four points $A$, $B$, $C$ and $D$
 is same after central projection from point $O$
 onto line $p$.
\end{geothm}

Código GCLCprover: 


Referência BiBTeX: 


Nome Utilizador: 
dim  130 85
point  A 20 20
point  C 40 20
point  D 55 20
point  B 90 20
point  O 80 70
line  ao A O
line  bo B O
line  co C O
line  do D O
point  A_1 32 30
point  B_1 84 50
line  ab1 A_1 B_1
intersec  C_1 ab1 co
intersec  D_1 ab1 do
cmark_b  A
cmark_b  B
cmark_b  C
cmark_b  D
cmark_r  O
cmark_rb  A_1
cmark_rb  C_1
cmark_rb  D_1
cmark_rb  B_1
drawsegment  A B
drawsegment  A O
drawsegment  B O
drawsegment  C O
drawsegment  D O
drawdashline  ab1



Nome do Teorema: 
point  A 20 20
point  C 40 20
point  D 55 20
point  B 90 20
point  O 80 70
line  ao A O
line  bo B O
line  co C O
line  do D O
online  A_1  A O
online  B_1  B O
line  ab1  A_1  B_1 
intersec  C_1  ab1  co
intersec  D_1  ab1  do
prove { equal { mult { sratio C A C B } { sratio D B D A } } { mult { sratio C_1 A_1 C_1 B_1 } { sratio D_1 B_1 D_1 A_1 } } }


Teorema: 


Descrição Breve: 
 Pappus Hexagon Theorem



Observações: \begin{geothm}
 Let $A$, $B$ and $C$ be three points on the
 line $p$, and let $A_1$, $B_1$ and $C_1$ be
 three points on the line $q$. Following intersections are created:
 $P = AB_1 \cap A_1B$, $Q = AC_1 \cap A_1C$ and
 $R = BC_1 \cap B_1C$. Show that points
 $P$, $Q$ and $R$ are collinear.
\end{geothm}

Código GCLC: 


Código GCLCprover: 


Referência BiBTeX: 
dim  110 90
point  A 40 10
point  B 90 10 
point  C 14.10 10
point  A_1 35.20 38.20
point  B_1 76.80 64.60
point  C_1 48.30 46.39
line  A_1B_1 A_1 B_1 
line  AB_1 A B_1
line  AC_1 A C_1
line  BA_1 B A_1
line  BC_1 B C_1
line  CA_1 C A_1
line  CB_1 C B_1
intersec  P AB_1 BA_1
intersec  Q AC_1 CA_1
intersec  S BC_1 CB_1
cmark_b  A
cmark_b  B
cmark_b  C
cmark_lt  A_1
cmark_rb  B_1
cmark_l  C_1
cmark_rb  P
cmark_r  Q
cmark_r  S
drawdashline  P S
drawsegment  A B_1
drawsegment  A Q
drawsegment  B A_1
drawsegment  B C_1
drawsegment  C Q
drawsegment  C B_1
drawline  A_1 C_1
drawline  A C



Nome Utilizador: 
point  A 40 10
point  B 90 10 
online  C  A B
point  A_1 35.20 38.20
point  B_1 76.80 64.60
online  C_1  A_1 B_1
line  A_1B_1 A_1 B_1 
line  AB_1 A B_1
line  AC_1  A C_1 
line  BA_1 B A_1
line  BC_1  B C_1 
line  CA_1  C  A_1
line  CB_1  C  B_1
intersec  P AB_1 BA_1
intersec  Q  AC_1  CA_1 
intersec  S  BC_1  CB_1 
prove { collinear P Q S }



Nome do Teorema: 


Teorema: gpreda1


Descrição Breve: \begin{geothm}
 Show that in any triangle, the orthocenter,
 the centroid and the center of circumscribed
 circle are collinear.
\end{geothm}

Observações: 


Código GCLC: 


Código GCLCprover: 
dim  110 110
point  A 20 20
point  B 90 20
point  C 70 80
midpoint  C_1 A B
midpoint  A_1 B C
line  at A A_1
line  ct C C_1
intersec  T at ct
med  a B C
med  c A B
intersec  O a c
line  ab A B
line  bc B C
foot  C_2 C ab
foot  A_2 A bc
line  ah A A_2
line  ch C C_2
intersec  H ah ch
line  oth O T
drawline  oth
drawdashsegment  A A_2
drawdashsegment  C C_2
drawdashsegment  O A_1
drawdashsegment  O C_1
drawdashsegment  A A_1
drawdashsegment  C C_1
cmark_rb  H
cmark_t  O
cmark_rb  T
cmark_b  A
cmark_b  B
cmark_t  C
cmark_b  C_1
cmark_b  C_2
cmark_rt  A_1
cmark_rt  A_2
drawsegment  A B
drawsegment  A C
drawsegment  C B



Referência BiBTeX: 
point  A 20 20
point  B 90 20
point  C 70 80
midpoint  C_1 A B
midpoint  A_1 B C
line  at A A_1
line  ct C C_1
intersec  T at ct
med  a B C
med  c A B
intersec  O a c
line  ab A B
line  bc B C
foot  C_2 C ab
foot  A_2 A bc
line  ah A A_2
line  ch C C_2
intersec  H ah ch
line  oth O T
prove { equal { sratio H T T O } { 2 } }


Nome Utilizador: 


Nome do Teorema: gpreda1


Teorema: 
 Nine Points Circle



Descrição Breve: 


Observações: 


Código GCLC: 
dim  110 110
point  A 20 20
point  B 90 20
point  C 80 100
line  ab A B
line  ac A C
line  bc B C
foot  D A bc
foot  E B ac
foot  F C ab
midpoint  G A B
midpoint  J B C
midpoint  I C A
line  ad A D
line  be B E
intersec  H ad be
midpoint  K A H
midpoint  L B H
midpoint  M C H
line  kj K J
line  gm G M
intersec  O kj gm
circle  k O G
drawcircle  k
drawdashline  kj
drawdashline  gm
cmark_b  A
cmark_b  B
cmark_t  C
cmark_r  D
cmark_l  E
cmark_b  F
cmark_b  G
cmark_lb  H
cmark_t  I
cmark_r  J
cmark_lb  M
cmark_b  K
cmark_t  L
cmark_t  O
drawsegment  A B
drawsegment  A C
drawsegment  B C
drawsegment  A D
drawsegment  B E
drawsegment  C F



Código GCLCprover: 
point  A 20 20
point  B 90 20
point  C 80 100
line  ab A B
line  ac A C
line  bc B C
foot  D A bc
foot  E B ac
foot  F C ab
midpoint  G A B
midpoint  J B C
midpoint  I C A
line  ad A D
line  be B E
intersec  H ad be
midpoint  K A H
midpoint  L B H
midpoint  M C H
line  kj K J
line  gm G M
intersec  O kj gm
circle  k O G
prove { perpendicular G M G E }



Referência BiBTeX: 


Nome Utilizador: gpreda1


Nome do Teorema: 
 The Construction of Isosceles Triangle



Teorema: \begin{geothm}
 Segment $AB$ is given. Let $m$ be median of $AB$
 and let circle $k$ be circle with center $A$ and
 half-diameter $B$. If $C$ is the intersection of
 circle $k$ and median $m$, show that triangle
 $ABC$ is isosceles.
\end{geothm}

Descrição Breve: 


Observações: 
dim  110 110
point  A 20 20
point  B 90 20
med  c A B
circle  k A B
intersec2  C C_1 k c
cmark_b  A
cmark_rb  B
cmark_rt  C
drawsegment  A B
drawsegment  A C
drawsegment  C B
drawline  c
drawcircle  k



Código GCLC: 
point  A 20 20
point  B 90 20
med  c A B
circle  k A B
intersec2  C C_1 k c
prove { equal { mult { angle B A C } { angle B A C } } { 3 } }



Código GCLCprover: 


Referência BiBTeX: 



\end{document}
