# 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.

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 $\mathrm{\angle }PCQ$.
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Step 1
Let ${\mathrm{\Gamma }}_{1}$ and ${\mathrm{\Gamma }}_{2}$ be our circles, ${\mathrm{\Gamma }}_{1}\cap {\mathrm{\Gamma }}_{2}=\left\{C,D\right\}$, TM and TK be common tangents to our circles, where $\left\{M,K\right\}\subset {\mathrm{\Gamma }}_{1}.$.
Also, let P and Q be placed on segments MT and KT respectively such that
CP and CQ are tangents to ${\mathrm{\Gamma }}_{2}$ and ${\mathrm{\Gamma }}_{1}$ respectively.
Now, let $TC\cap {\mathrm{\Gamma }}_{1}=\left\{C,N\right\}$ and let f be a homothety with the center T such that $f\left({\mathrm{\Gamma }}_{2}\right)={\mathrm{\Gamma }}_{1}$.
Step 2
Thus, $f\left(\left\{C\right\}\right)=\left\{N\right\}$ and f(CP) is a tangent to the circle ${\mathrm{\Gamma }}_{1}$ in the point N, which is parallel to CP and let this tangent intersects CQ in the point R.
Let $CP\cap {\mathrm{\Gamma }}_{1}=\left\{L,C\right\}$.
Thus, since NR||LP, we obtain $\measuredangle PCT=\measuredangle NCL=\measuredangle RNC=\measuredangle RCN=\measuredangle QCT$ and we are done