Balti Tee 2019 · Ülesanne 14

Let $K$ be the midpoint of segment $BH$, $S$ be the intersection point of $AK$ and $BP$, $T$ be the intersection point of $CK$ and $BQ$. Then $SK$ and $KT$ is one third of the corresponding medians and $ST$ is parallel to $AC$.

The triangles $ABH$ and $BCH$ are similar. From this similarity and properties of inscribed angles we have

$$\angle KAB = \angle NBC = \angle QAC.$$

Hence $\angle BAC = \angle KAR$. But also $\angle BAC = \angle BPC = \angle BPR$ as inscribed angles. Therefore quadrilateral $SAPR$ is cyclic and $\angle ARS = \angle APS = \angle APB = \angle AQB$. So, $RS$ is parallel to $BT$. By analogous reasoning $RT$ is parallel to $BS$. Hence $BSRT$ is parallelogram.

Diagonal $BR$ of this parallelogram splits diagonal $ST$ on 2 equal parts, therefore it also splits the segment $MN$ which is parallel to $ST$ on 2 equal parts, QED.