Question 73
The isosceles triangle $ABC$ has $AB = AC$ and $\angle CAB = 30^\circ$. Let $F \in [AB]$, $E \in [AC]$, $P$ on the line $AB$, and $M \in [BC]$ be points such that $\angle FCA = 15^\circ$, $\angle FEA = 45^\circ$, $CP \perp CF$ and $[PM]$ is the bisector of $\angle EPC$. Prove that $E$ is the midpoint of the segment $AC$ and $M$ is the midpoint of the segment $BC$.