Question 119
Given a Nonisceles Acute-Angled Triangle $ABC$. Let $O$ Be its CircumCenter, Let $AD$ Be its altitude, and let $M$ Be the Midpoint of Side $BC$. The Lines Passing Through $O$, perpendicular to $AB$ and $AC$, intersect $AD$ at points $P$ and $Q$, respectively. Let $S$ Be the CircumCenter of Triangle $OPQ$. prove that $\angle BAS = \angle CAM$. through the center $O$ circumference described near an unequal sharp triangle $ABC$, carried out straight, perpendicular to the sides $AB$ and $AC$. These lines cross the height $AD$ triangle $ABC$ at points $P$ and $Q$. Dot $M$ - the middle of the side $BC$, whereas $S$ - the center of the circle described near the triangle $OPQ$. Prove that $\angle BAS = \angle CAM$.