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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 !

2 Answers 2

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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)

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This follows becuse $DP$ and $FP$ have the same length, being segments hitting the sides from the diagonal at equal angles.

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    @Iuli: There was a comment from you here earlier, now deleted, asking me to explain this in mathematical notation: $DP\perp FP\land CD\perp CB\rightarrow \angle PDC=\angle PFB$.2012-08-09