Problem on an isosceles right triangle, involving similarity and congruence

Question

Given that $ABC$ is an isosceles right angled triangle with angle $\widehat{ACB}=90$ degrees. $D$ is the midpoint of $BC$, $CE$ is perpendicular to $AD$, intersecting $AB$ and $AD$ at $E$ and $F$ respectively. Prove that angle $\widehat{CDF}$ equals angle $\widehat{BDE}$.

I've tried some angle chasing, but that's pretty much it. I see that triangle $ACD$ is similar to some others, but I've not been able to use that properly.

Answer 1

Proof without words. Well, I suppose I need some words for the character limit.