50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

Let $O$ be the circumcenter of $ABC$. Let our perpendicular bisectors meet sides $AB$ and $BC$ at points $K$ and $L$ respectively. Then $KL$ is a midline of triangle $ABC$. Therefore, $KL\parallel AC$ and $2KL=AC$. Thus, $MKLN$ is parallelogram, because $KL\parallel MN$ and $KL=MN$. Due to the fact, that $PM\perp AC$ and $QN\perp AC$, we have $PN\parallel QN$. Moreover, $PM\parallel QN$ and $KM\parallel LN$, thus, $\angle PMK=\angle QNL$ (Fig.10).

$\angle AKP=\angle AMP=90^{\circ }$, thus $P,K,M$ and $A$ lie on the same circle with diameter $PA$. By analogy, we obtain that $Q,N,C$ and $L$ lie on the same circle with diameter $CQ$, points $O,K,B$ and $L$ belong to the circle with diameter $BO$. From these observations, it follows $\angle PAK=\angle PMK=\angle QNL=\angle QCL$.

$P$ and $Q$ belong to perpendicular bisectors of the segments $AB$ and $BC$ respectively, thus $APB$ and $BQC$ are isosceles triangles. From this, we obtain $\angle PBA=\angle PAB=\angle QCB=\angle QBC$. Hence, $\angle PBQ=\angle KBL=180^{\circ }-\angle KOL=180^{\circ }-\angle POQ$, in other words, points $P,B,Q,O$ are cyclic, which provides desired result.