# Prove that $CE=AB$

Suppose $ABC$ is an acute-angled triangle with $AB<AC$.Let $M$ be the midpoint of $BC$. Suppose $P$ is a point on side $AB$ such that, if $PC$ intersects the median $AM$ at E, then $AP=PE$.Prove that $AB=CE$.

I don't know how to start. Please give me an idea. Getting no fruitful thoughts ,I started using barycentric Coordinates.But the calculations seemed very tough and I failed to proceed. Please give me any idea to start

• I successfully bashed the problem with barycentric coordinates within 1 page!! – Sufaid Saleel Sep 11 '18 at 2:17

Hint:   write Menelaus' theorem for triangle $\,\triangle PBC\,$ and transversal $\,AM\,$.