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Geometry Difficulty 4.5 AIME Prove it Singapore

In a convex hexagon $ABCDEF$ , $AB$ is parallel to $DE$ , $BC$ is parallel to $EF$ and $CD$ is parallel to $FA$ . Prove that the triangles $ACE$ and $BDF$ have the same area.

Solution

Let the coordinates of $A$ be $(0,0)$ , $B$ be $(1,0)$ , $D$ be $(c,b)$ , $E$ be $(a,b)$ , $F$ be $(e,f)$ and $C$ be $(x,y)$ . Then $AB \parallel ED$ . For $AF \parallel CD$ , we need $(b-y)/(c-x) = f/e$ . For $EF \parallel CB$ we need $(x-1)/y = (a-e)/(b-f)$ . From these equations, we get
$xb - ay = b - f + cf - be. \quad (*)$
Thus the proof is now complete since the following is true by $(*)$ .
$[ACE] = [BDF] \Leftrightarrow \begin{vmatrix} 0 & 0 & 1 \\ x & y & 1 \\ a & b & 1 \end{vmatrix} = \begin{vmatrix} 1 & 0 & 1 \\ c & b & 1 \\ e & f & 1 \end{vmatrix} \Leftrightarrow xb - ay = b - f + cf - be.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.