euclidean-geometrygeometry
AEBCD is a convex pentagon in which ABCD is a square. The Diagonals ED
and AB meet at F, while diagonals EC and AB meet at G. Prove that the sum of the areas of triangles EAF and EBG equal the area of triangle DFG?
Some things I noticed is that triangle EAF and DFG have a common vertical angle and that ABCD contains 2 right triangles. I don't know if that's helpful.
Your approach looks fine to me. Indeed, the most natural way to show that a line bisects the area of a square is to show that it passes through the center of that square. Here this follows from the fact that $O$ is the midpoint of the arc $AB$ of the circumcircle of $\triangle APB$ and the fact that the inner angle bisector in a triangle passes through the midpoint of the opposite arc of the circumcircle.
If one among the angles at $A$, $B$, $C$, $D$ is a right angle, then it is easy to prove they are all right angles and $ABCD$ is a square. Suppose then none of them is a right angle: at least one of them ($\angle GAH$, for instance) must then be obtuse. I'll show that this leads to a contradiction.
Let $N$ be the foot of the perpendicular from $G$ to line $AH$. As $\angle GAH>90°$ then $AH<NH$.
Let $M$ be the foot of the perpendicular from $E$ to line $BH$: we have then $EM\le EB$. But triangles $EMH$ and $HNG$ are congruent (because they have $\angle NGH=\angle MHE=\pi/2-\alpha$, $\angle NHG=\angle MEH=\alpha$ and $GH=HE$), thus: $$ AH < NH = EM \le EB $$ which is in contradiction with the hypothesis $AH=EB$.
Best Answer
Let [.] denote triangle areas. Establish the area ratios below,
$$\frac{[AEF]}{[EFG]}=\frac{AF}{FG},\>\>\>\>\>\frac{[EGB]}{[EFG]}=\frac{GB}{FG}, \>\>\>\>\>\frac{[FGD]}{[EFG]}=\frac{DF}{EF}$$
because all three pairs share same height. Then,
$$\frac{[AEF]+[EGB]}{[FGD]} =\frac{\left(\frac{AF}{FG}+\frac{GB}{FG}\right)[EFG]}{\frac{DF}{EF}[EFG]} =\frac{\frac{AB-FG}{FG}}{\frac{DE-EF}{EF}} =\frac{\frac{AB}{FG}-1}{\frac{DE}{EF}-1} $$
Given that triangles EFG and EDC are similar, we have $\frac{DC}{FG}=\frac{AB}{FG}=\frac{DE}{EF}$ and
$$\frac{[AEF]+[EGB]}{[FGD]} = 1$$