A circle $\omega$ is inscribed in the acute-angled triangle $\vartriangle ABC$, which touches the side $BC$ at the point $K$. On the lines $AB$ and $AC$, the points $P$ and $Q$, respectively, are chosen so that $PK \perp AC$ and $QK \perp AB$. Denote by $M$ and $N$ the points of intersection of $KP$ and $KQ$ with the circle $\omega$. Prove that if $MN \parallel PQ$, then $\vartriangle ABC$ is isosceles. (S. Slobodyanyuk)