It is proved that the angle EBF of the intersection point B is 90 degrees, which makes BF equal to AD and connects CF.

So, ∠EBF=90

And because in the isosceles triangle ABC, ∠A=∠B=45.

Therefore, < ∠CBF=45.

Because, AC = AB and AD = BF.

So triangle CAD is equal to triangle CBF (corner edge)

So ∠ACF=∠BCF, CD=CF because ∠ACD+∠DCB=90.

So ∠BCF+∠DCB=90, which means ∠DCF=90.

Because ∠ DCE = 90

So ∠ECF=45 connects E and F.

Because CD=CF, ∠DCE=∠ECF=45, CE=CE.

So all triangles DCE are equal to triangle ECF (corner edge)

So DE=DF, so in the right triangle EBF? +BF？ =EF？ So de? =AD？ +BE？