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.