Selection and Training Session

Let $M$, $N$ and $R$ be the midpoints of the sides $AB$, $AC$, $BC$, respectively. Let $X$ denote the foot of the perpendicular from $O$ on $AH$. We claim that $X$ is the point of intersection of ${\omega }_1$ and ${\omega }_2$.



Since $O$ is the center of the circumcircle of the triangle $ABC$, we have $OR\perp BC$, $ON\perp AC$ and $OM\perp AB$. So $O$, $M$, $A$, $N$ and $X$ lie on the same circle ${\omega }_1$, and $OA$ is the diameter of ${\omega }_1$.

Show that $PR=HQ$. Indeed, $$PR=BR-BP=\frac{1}{2}(BC-BH)=\frac{1}{2}HC=HQ.$$ From $PR=HQ$ and $OX\parallel BC$ it follows that $POXQ$ is an isosceles trapezium. Therefore $X$ belongs to ${\omega }_2$.