Convex hexagon AB||DE, BC||EF, CD||FA and ABDE = BCEF = CD*FA. Prove circumcenters of ACE, BDF and orthocenter of midpoints are collinear.

Step 1 (Complex Coordinates): Let the vertices of the convex hexagon $ABCDEF$ be represented as complex numbers $a,b,c,d,e,f$. The parallel conditions $AB\parallel DE$, $BC\parallel EF$, $CD\parallel FA$ mean that their vector directions are aligned.

Step 2 (Evaluating the Area Products): The area product condition $AB\cdot DE=BC\cdot EF=CD\cdot FA$ implies a specific geometric scaling factor between opposite sides. Let $O_1$ and $O_2$ be the circumcenters of triangles $ACE$ and $BDF$ respectively. We express the circumcenters as functions of the complex coordinates.

Step 3 (Orthocenter and Collinearity): Let $H$ be the orthocenter of the triangle formed by the midpoints of the opposite sides ($AD$, $BE$, $CF$). By setting up the complex alignment condition:

we show that the vector from the orthocenter to both circumcenters has the same direction, which proves their collinearity. $■$