Two circles intersect at $C$ and $D$, and their common tangents intersect at $T$. $CP$ and $CQ$ are the tangents at $C$ to the two circles; prove that $CT$ bisects $\angle PCQ$.
I am aware of some basic theorems on the centres of similitude/homothety, but I'm not sure how to apply them to compute angles. Most of the other problems here have been about computing ratios of lengths.
Let $\Gamma_1$ and $\Gamma_2$ be our circles, $\Gamma_1\cap\Gamma_2=\{C,D\}$,
$TM$ and $TK$ be common tangents to our circles, where $\{M,K\}\subset\Gamma_1.$
Also, let $P$ and $Q$ be placed on segments $MT$ and $KT$ respectively such that
$CP$ and $CQ$ are tangents to $\Gamma_2$ and $\Gamma_1$ respectively.
Now, let $TC\cap\Gamma_1=\{C,N\}$ and let $f$ be a homothety with the center $T$ such that $f(\Gamma_2)=\Gamma_1$.
Thus, $f(\{C\})=\{N\}$ and $f(CP)$ is a tangent to the circle $\Gamma_1$ in the point $N$,
which is parallel to $CP$ and let this tangent intersects $CQ$ in the point $R$.
Let $CP\cap \Gamma_1=\{L,C\}$.
Thus, since $NR||LP,$ we obtain $$\measuredangle PCT=\measuredangle NCL=\measuredangle RNC=\measuredangle RCN=\measuredangle QCT$$ and we are done!