Problem

Source: 38th Brazilian MO (2016) - First Day, Problem 1

Tags: geometry, incenter, Parallel Lines, Brazilian Math Olympiad 2016



Let $ABC$ be a triangle. $r$ and $s$ are the angle bisectors of $\angle ABC$ and $\angle BCA$, respectively. The points $E$ in $r$ and $D$ in $s$ are such that $AD \| BE$ and $AE \| CD$. The lines $BD$ and $CE$ cut each other at $F$. $I$ is the incenter of $ABC$. Show that if $A,F,I$ are collinear, then $AB=AC$.