inside equilateral triangle $ABC$, $BD=CE$, $AE \cap CD = F$. Find maximum area of $BDFE$
I could prove through extremely tedious computation that $S_{BDFE}$ maximizes when $D$ and $E$ are the midpoints of $AB,BC$, but I would like to see if there is a more elegant solution.
Draw a third line from $B$ to cut $AC$ at $G$, so that $AG = BD = CE$.
By symmetry, the three lines $AE$, $BG$, $CD$ cut an equilateral triangle inside the triangle (with one vertex $F$), and leave an outer ring which has an area equals to three times that of $BDFE$.
To maximise the area of the outer ring and therefore $BDFE$, the area of the inner equilateral triangle should be minimised. This happens when the perpendicular distance from $AE$ to the common centre of the equilateral triangles is minimised to $0$.