Question 181
Given are circles $\Gamma_{1}$ with center $A$ and $\Gamma_{2}$ with center $B$, where $A$ lies on $\Gamma_{2}$. On $\Gamma_{2}$ there is furthermore a variable point $P$, not on $A B$. A line through $P$ that touches $\Gamma_{1}$ at $S$ intersects $\Gamma_{2}$ again at $Q$, where $P$ and $Q$ lie on the same side of $A B$. Another line through $Q$ touches $\Gamma_{1}$ at $T$. Let furthermore $M$ be the foot of the perpendicular from $P$ to $A B$. Let $N$ be the intersection of $A Q$ and $M T$. Prove that $N$ lies on a line that is independent of the position of $P$ on $\Gamma_{2}$.