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
On
Construct the //gm BECF.
From (1) AEM is a straight line; (2) E is a vertex of that //gm; (3) M is the midpoint of one of its diagonals; and (4) F is a vertex of that //gm and FM is a straight line, we can say that AEMF is a straight line.
The required result follows because all the green marked angles are equal.
Hint: write Menelaus' theorem for triangle $\,\triangle PBC\,$ and transversal $\,AM\,$.