\newcommand\proofstep[2]{#1 & \mbox{#2} \\ }

\begin{eqnarray}
\proofstep{\left(\left( \frac{\overrightarrow{AF}}{\overrightarrow{FB}} \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{EA}}\right) =  1 }{by statement}
\proofstep{\left(\left(\left( -1  \cdot  \frac{\overrightarrow{AF}}{\overrightarrow{BF}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{EA}}\right) =  1 }{by Lemma 2.1}
\proofstep{\left(\left(\left( -1  \cdot  \frac{ S_{APC}}{ S_{BPC}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{EA}}\right) =  1 }{by Lemma 2.4}
\proofstep{\left(\left(\left( -1  \cdot  \frac{ S_{APC}}{ S_{BPC}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot \left( -1  \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{AE}}\right)\right) =  1 }{by Lemma 2.1}
\proofstep{\left(\left(\left(\left( -1  \cdot  \frac{ S_{APC}}{ S_{BPC}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  -1 \right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{AE}}\right) =  1 }{by left association rule}
\proofstep{\left(\left( -1  \cdot \left(\left( -1  \cdot  \frac{ S_{APC}}{ S_{BPC}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right)\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{AE}}\right) =  1 }{by commutativity rule}
\proofstep{\left(\left(\left( -1  \cdot \left( -1  \cdot  \frac{ S_{APC}}{ S_{BPC}}\right)\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{AE}}\right) =  1 }{by left association rule}
\proofstep{\left(\left(\left(\left( -1  \cdot  -1 \right) \cdot  \frac{ S_{APC}}{ S_{BPC}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{AE}}\right) =  1 }{by left association rule}
\proofstep{\left(\left(\left( 1  \cdot  \frac{ S_{APC}}{ S_{BPC}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{AE}}\right) =  1 }{by multiplication of constants}
\proofstep{\left(\left( \frac{ S_{APC}}{ S_{BPC}} \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\overrightarrow{CE}}{\overrightarrow{AE}}\right) =  1 }{by multiplication by 1}
\proofstep{\left(\left( \frac{ S_{APC}}{ S_{BPC}} \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{ S_{CPB}}{ S_{APB}}\right) =  1 }{by Lemma 2.4}
\proofstep{\left(\left( \frac{ S_{APC}}{ S_{BPC}} \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{DC}}\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{ S_{APB}}\right) =  1 }{by Lemma 2.2}
\proofstep{\left(\left( \frac{ S_{APC}}{ S_{BPC}} \cdot \left( -1  \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{CD}}\right)\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{ S_{APB}}\right) =  1 }{by Lemma 2.1}
\proofstep{\left(\left(\left( \frac{ S_{APC}}{ S_{BPC}} \cdot  -1 \right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{CD}}\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{ S_{APB}}\right) =  1 }{by left association rule}
\proofstep{\left(\left(\left( -1  \cdot  \frac{ S_{APC}}{ S_{BPC}}\right) \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{CD}}\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{ S_{APB}}\right) =  1 }{by commutativity rule}
\proofstep{\left(\left( \frac{\left( -1  \cdot  S_{APC}\right)}{ S_{BPC}} \cdot  \frac{\overrightarrow{BD}}{\overrightarrow{CD}}\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{ S_{APB}}\right) =  1 }{by mult simplification rules}
\proofstep{\left(\left( \frac{\left( -1  \cdot  S_{APC}\right)}{ S_{BPC}} \cdot  \frac{ S_{BPA}}{ S_{CPA}}\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{ S_{APB}}\right) =  1 }{by Lemma 2.4}
\proofstep{\left(\left( \frac{\left( -1  \cdot  S_{APC}\right)}{ S_{BPC}} \cdot  \frac{ S_{BPA}}{\left( -1  \cdot  S_{APC}\right)}\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{ S_{APB}}\right) =  1 }{by Lemma 2.2}
\proofstep{\left(\left( \frac{\left( -1  \cdot  S_{APC}\right)}{ S_{BPC}} \cdot  \frac{ S_{BPA}}{\left( -1  \cdot  S_{APC}\right)}\right) \cdot  \frac{\left( -1  \cdot  S_{BPC}\right)}{\left( -1  \cdot  S_{BPA}\right)}\right) =  1 }{by Lemma 2.2}
\proofstep{\left(\left( \frac{\left( -1  \cdot  S_{APC}\right)}{ S_{BPC}} \cdot  \frac{ S_{BPA}}{\left( -1  \cdot  S_{APC}\right)}\right) \cdot  \frac{\left( 1  \cdot  S_{BPC}\right)}{\left( 1  \cdot  S_{BPA}\right)}\right) =  1 }{by cancellation rule}
\proofstep{\left(\left( \frac{\left( -1  \cdot  S_{APC}\right)}{ S_{BPC}} \cdot  \frac{ S_{BPA}}{\left( -1  \cdot  S_{APC}\right)}\right) \cdot  \frac{ S_{BPC}}{\left( 1  \cdot  S_{BPA}\right)}\right) =  1 }{by multiplication by 1}
\proofstep{\left(\left( \frac{\left( -1  \cdot  S_{APC}\right)}{ S_{BPC}} \cdot  \frac{ S_{BPA}}{\left( -1  \cdot  S_{APC}\right)}\right) \cdot  \frac{ S_{BPC}}{ S_{BPA}}\right) =  1 }{by multiplication by 1}
\proofstep{\left( \frac{\left(\left( -1  \cdot  S_{APC}\right) \cdot  S_{BPA}\right)}{\left( S_{BPC} \cdot \left( -1  \cdot  S_{APC}\right)\right)} \cdot  \frac{ S_{BPC}}{ S_{BPA}}\right) =  1 }{by mult simplification rules}
\proofstep{\left( \frac{\left( 1  \cdot  S_{BPA}\right)}{\left( S_{BPC} \cdot  1 \right)} \cdot  \frac{ S_{BPC}}{ S_{BPA}}\right) =  1 }{by cancellation rule}
\proofstep{\left( \frac{ S_{BPA}}{\left( S_{BPC} \cdot  1 \right)} \cdot  \frac{ S_{BPC}}{ S_{BPA}}\right) =  1 }{by multiplication by 1}
\proofstep{\left( \frac{ S_{BPA}}{\left( 1  \cdot  S_{BPC}\right)} \cdot  \frac{ S_{BPC}}{ S_{BPA}}\right) =  1 }{by commutativity rule}
\proofstep{\left( \frac{ S_{BPA}}{ S_{BPC}} \cdot  \frac{ S_{BPC}}{ S_{BPA}}\right) =  1 }{by multiplication by 1}
\proofstep{ \frac{\left( S_{BPA} \cdot  S_{BPC}\right)}{\left( S_{BPC} \cdot  S_{BPA}\right)} =  1 }{by mult simplification rules}
\proofstep{ \frac{\left( 1  \cdot  S_{BPC}\right)}{\left( S_{BPC} \cdot  1 \right)} =  1 }{by cancellation rule}
\proofstep{ \frac{\left( 1  \cdot  1 \right)}{\left( 1  \cdot  1 \right)} =  1 }{by cancellation rule}
\proofstep{ 1  =  1 }{by cancellation rule}
\end{eqnarray}

 \centerline{Q.E.D.}

