Saudi Arabian Mathematical Competitions

Let $R$ be the intersection point of lines $PQ$ and $BC$, and $S$ the intersection point of lines $MT$ and $BC$.

Because circle $\omega$ is tangent to both sides $AB$ and $AC$ at $P$ and $Q$, we have $AP = AQ$. Applying Menelaus' theorem to triangle $ABC$ and line $PQ$ we obtain
$$\frac{\overline{BR}}{\overline{RC}} = -\frac{BP}{PA} \cdot \frac{AQ}{QC} = -\frac{BP}{QC}.$$
Because $M$ is the midpoint of arc $\widehat{BC}$ of circle $\Omega$ containing $T$, line $MT$ is the external bisector of angle $\angle BTC$. We deduce that
$$\frac{\overline{BS}}{\overline{SC}} = -\frac{TB}{TC}$$
To prove that $S = R$, we have to prove that $\frac{\overline{BS}}{\overline{SC}} = \frac{\overline{BR}}{\overline{RC}}$, which is equivalent to proving that
$$\frac{BP}{QC} = \frac{TB}{TC}.$$
Let $B'$ and $C'$ be the second intersection points of lines $TB$ and $TC$ with circle $\omega$, respectively. Because circles $\Omega$ and $\omega$ are tangent at $T$, segments $BC$ and $B'C'$ are parallel, and therefore
$$\frac{TB}{TC} = \frac{BB'}{CC'}.$$
Because lines $AB$ and $AC$ are tangent to $\omega$ at $P$ and $Q$, respectively, we have from the power of points $B$ and $C$ with respect to circle $\omega$
$$\left(\frac{BP}{CQ}\right)^2 = \frac{BT \cdot BB'}{CT \cdot CC'} = \left(\frac{BT}{CT}\right)^2.$$