ABCDE is a convex pentagon with ABCD a square. F and G are on AB. Prove EAF + EBG = DFG in areas

Question

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.

Answer 1

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

$$\frac{[AEF]+[EGB]}{[FGD]} = 1$$

Answer 2

Through E, draw EXY //AD cutting AB at X and CD at Y.

DFG = (DFX) + [DXG]

= (EAF) + [DXG]

= (EAF) + [XGC]

= (EAF) + [EGB] , because triangles with “equal altitude and equal base” yields equal area.