Given two angles, exact location of $E$ and $F$ in a square $ABCD$ and diagonal $BD$ intersects $AE$ at $P$. What is the value of $\angle PFC$?

Question

$ABCD$ is a square. $E$ and $F$ are points on $BC$ and $CD$ such that $\angle EAF$ = 45$^\circ$ and $\angle EAB$ = 15$^\circ$. Diagonal of the square $BD$ intersects $AE$ at the point $P$. Than point $P$ and $C$ are connected. What is the value of $\angle PFC$ in degrees?

This problem appeared in the math-contest of Bangladesh. I couldn't figure it out because I didn't know which segments of the following square should be attached and which new lines should be drawn through for further a easy solution.

I will be very gladful if anyone shows me how to find out $\angle PFC$ or help me learn some better strategy for that or similar problem of mensuration.

Answer 1

Since $$\measuredangle FAP=\measuredangle ADP=45^{\circ},$$ we see that $APFD$ is cyclic,

which says $FP\perp AE$, which says $PECF$ is cyclic, which gives $$\measuredangle PFC=\measuredangle AEB=75^{\circ}.$$

Answer 2

Hint:

$$\angle PAF = \angle EAF = 45^{\circ} = \angle BDC = \angle PDF,$$

so $ADFP$ is cyclic. Thus $PF \perp AE$. Can you go from here?