TITLE: A geometry problem - measure of an $\angle$
QUESTION [5 upvotes]: We have a square and the following information: 
1) $E \in [AB]$, $E$ an arbitrary point  
2) $[AC] \cap [DE]= \{P\}$ and 
3)$FP \perp ED$, where $F \in BC$ . 
We have to prove that the measure of the angle $\angle EDF = 45^{\circ}$. 
 
Thanks a lot !

REPLY [2 votes]: It's easy to see that $PFCD$ is a cyclic quadrilateral, $\angle DPF$ + $\angle FCD = 180^{\circ}$. Therefore, we have 
$$\angle EDF = \angle PDF = \angle PCF=\angle ACB = 45^{\circ}$$
Q.E.D. (a very simple problem)