2015 OIM Problems/Problem 2

Problem

A line $r$ contains the points $A$, $B$, $C$ and $D$ in that order. Let $P$ be a point outside of $r$ such that $\angle APB = \angle CPD$. Prove that the bisector of $\angle APD$ cuts $r$ at a point $G$ such that

\[\frac{1}{GA}+\frac{1}{GC}=\frac{1}{GB}+\frac{1}{GD}\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

OIM Problems and Solutions