EDBC is a trapezoid
We know AC=8 and since the sides of an isosceles right triangle are in ratio
[tex] 1 : 1 : \sqrt 2[/tex]
we conclude
[tex]AB = BC = AC/\sqrt{2} = 8/sqrt{2} = 4 \sqrt 2[/tex]
We have a square so AD=ED=EF=4.
DB is the height of the trapezoid,
[tex]DB = AB - AD = 4 \sqrt 2 - 4[/tex]
So the area of the trapezoid is
[tex] t = \frac 1 2 (b_1+b_2)h = \frac 1 2 (4 \sqrt{2} + 4)(4 \sqrt 2 - 4) = 8(\sqrt 2 + 1)(\sqrt 2 -1) = 2(2 - 1) = 8[/tex]
AFECB is the sum of two isosceles right triangles AFE + ABC so has area
[tex]p = \frac 1 2 (4)^2 + \frac 1 2 (4 \sqrt{2})^2 = \frac 1 2 (16 + 32) =24[/tex]
That's a ratio of 8:24:8 or 1:3
That must mean that ADE and AFE each have area 8 as well.
Answer: 1/3
