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