Question 105
Let $\omega$ Be the Circle of a Triangle $ABC$. a point $K$ Is Chosen on the Bisector Line of Angle $BAC$ so that $K$ lies inside triangle $ABC$. line $CK$ intersects $\omega$ at points $C$ and $M$. a circle $\Omega$ Touches Line $CM$ at $K$, and intersects segment $AB$ at points $A$ and $P$. Circles $\Omega$ and $\omega$ intersect at points $A$ and $Q$. Prove that points $P$, $Q$, and $M$ Are Collinear. inside the triangle $ABC$ Point taken $K$lying on the bisector of the angle $BAC$. Straight $CK$ crosses a circle a second time $\omega$described near the triangle $ABC$, at the point $M$. Circle $\Omega$ Passes through a point $A$, touches the straight line $CM$ at the point $K$ and crosses the segment a second time $AB$ at the point $P$, a circle $\omega$ - at the point $Q$. Prove that points $P$, $Q$ and $M$ lie on one straight line.