Let $ABC$ be a triangle for which the shortest side is $AC$. Its inscribed circle with center $I$ touches sides $AB$ and $BC$ in points $D$ and $E$ respectively. Point $M$ is the midpoint of $AC$. Points $F$ and $G$ lie on sides $BC$ and $AB$ respectively so that $FC=CA=AG$. The line through $I$ perpendicular to $MI$ intersects the line segments $AF$ and $CG$ in $P$ and $Q$ respectively. Prove that $AB=BC\Leftrightarrow PD=QE$.