Let $M$ be a point on the diameter $AB$ of a semicircle $\Gamma$. The perpendicular at $M$ meets the semicircle $\Gamma$ at $P$. A circle inside $\Gamma$. touches $\Gamma$. and is tangent to $PM$ at $Q$ and $AM$ at $R$. Prove that $P B = RB$.
Source: 2003 Singapore TST 2.2
Tags: geometry, equal segments, semicircle
Let $M$ be a point on the diameter $AB$ of a semicircle $\Gamma$. The perpendicular at $M$ meets the semicircle $\Gamma$ at $P$. A circle inside $\Gamma$. touches $\Gamma$. and is tangent to $PM$ at $Q$ and $AM$ at $R$. Prove that $P B = RB$.