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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 !

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up vote 2 down vote accepted

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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it is not clearly. and how can I prove that the segments $DP$ and $FP$ are hitting the sides from the diagonal at equal angles? Thank ! – Iuli Aug 8 '12 at 14:57
How do you mean "it is not clearly"? Do you mean the answer isn't clear, or that it's clearly wrong? The segments hit the sides at equal angles because both the segments and the sides are at right angles to each other. – joriki Aug 8 '12 at 14:59
@Iuli: There was a comment from you here earlier, now deleted, asking me to explain this in mathematical notation: $DP\perp FP\land CD\perp CB\rightarrow \angle PDC=\angle PFB$. – joriki Aug 9 '12 at 6:16

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