Mongolian Mathematical Olympiad

Let's prove that all possible lines $PQ$ pass through the point $I$.



In order to prove this we need to prove that $\angle PIM+\angle QIM=180^{\circ }$. It is easy see $MI=MB=MC$. It follows from $\angle IBM=\angle BIM=\frac{\alpha +\beta }{2}$, $\angle ICM=\angle CIM=\frac{\alpha +\gamma }{2}$. Since $\angle BPM=\angle BCM=\angle DBM$, by AA criterion $△BPM\sim △DBM$ and we get $\frac{PM}{BM}=\frac{BM}{DM}$. Considering that $BM=IM$ from where follows $\frac{PM}{IM}=\frac{IM}{DM}$.

Triangles $△PMI$, $△IMD$ have the common angle $\angle PMI=\angle IMD$ so we have $△PMI\sim △IMD$. Therefore we conclude that $\angle PIM=\angle IMD$.

Similarly, $△CQM\sim △ECM$ implies $\frac{QM}{CM}=\frac{CM}{EM}$ and $△QMI\sim △IME$ consequently we get $\angle QIM=\angle IEM$. Since the points $I$, $E$, $M$, $D$ lie on the circle $\gamma$, $\angle IDM+\angle IEM=180^{\circ }$ and thus we have proved that $\angle PIM+\angle QIM=180^{\circ }$.