Ratio of 2 Triangles

By symmetry, $DF = BE = x$ . Let $AE = EF = FA = y$

Also, we observe that $EC = CF = a - x$

In $\triangle ABE$ :

$y^{2} = a^{2} + x^{2}$ ... (i)

In $\triangle ECF$ :

$y^{2} = (a-x)^{2} + (a-x)^{2}$

Using (i):

$a^{2} + x^{2} = 2(a-x)^{2}$

$a^{2} + x^{2} - 4ax = 0$

$a^{2} + x^{2} - 2ax = 2ax$

$(a-x)^{2} = 2ax$ ... (ii)

$\frac{Ar\triangle ABE}{Ar\triangle ECF}$ = $\frac{0.5ax}{0.5(a-x)^{2}}$

Using (ii):

$\frac{Ar\triangle ABE}{Ar\triangle ECF}$ = $\frac{ax}{2ax}$ = $1:2$

Let (wlog) , FC = EC =2, then FE = 2√2, [FEC] = (2√2)(√2)/2 = 2. [ABE] = (2√2)(2√2)cos 15 . sin 15 /2 = 2 cos 30 = 1. [ABE]/[FEC] = 1/2