Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it

The circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at $D$ and $P$ . The common tangent of the two circles closest to point $D$ touches $\Gamma_{1}$ at $A$ and $\Gamma_{2}$ at $B$ . The line $A D$ intersects $\Gamma_{2}$ again at $C$ . Let $M$ be the midpoint of segment $B C$ .
Prove that $\angle D P M=\angle B D C$ .

保留了源文本的换行和格式，直接输出了翻译结果。

Solution

Let $S$ be the intersection of $PD$ and $AB$ . Then $S$ lies on the radical axis of the two circles, and thus $|SA|=|SB|$ . Therefore, $PS$ is a median in triangle $PAB$ .
By the tangent-secant angle theorem on $\Gamma_{1}$ with chord $AP$ , we have $\angle BAP=180^{\circ}-\angle ADP=\angle CDP=\angle CBP$ . By the tangent-secant angle theorem on $\Gamma_{2}$ with chord $BP$ , we have $\angle ABP=\angle BCP$ . Thus, $\triangle PAB \sim \triangle PBC$ (AA). Since $PM$ is a median in triangle $PBC$ , we have $\angle SPB=\angle MPC$ . Therefore,
$\angle DPM=\angle DPB+\angle BPM=\angle SPB+\angle BPM=\angle MPC+\angle BPM=\angle BPC=\angle BDC$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.