MathNet 2024-part-03 — 168 problems (EN–VI) Question 115

Let $ABC$ be an isosceles triangle with $AB = BC$. Let $M$ and $N$ be midpoints of $AC$ and $BM$, respectively. $P$ is the foot of the altitude from $M$ to $AN$ of triangle $AMN$. Prove that triangles $APM$ and $CPB$ are similar.
(Khulan Tumenbayar)