Let $\triangle ABC$ be a triangle and $r$ and $s$ be 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 $AB=AC$.
What I thought:
Since $AE || CD $ and $AD || BE $, so, $\frac {FC}{CE} $ = $\frac {FI}{IA} $ = $\frac {FB}{BD} $
Since $AE\parallel IC$ and $AD\parallel IB$, it follows that $\frac{CE}{EF}=\frac{IA}{AF}=\frac{BD}{DF}$. Hence $DE\parallel BC$.
$\angle ADE=\angle DEB=\angle CBE=\angle ABE$
Hence a circle can be circumscribed about trapezoid $ADBE$. It follows that $AB=DE$.
Similarly $AC=DE$.