Problem 10.1. A circle $k$ is tangent to the arms of an acute angle $A O B$ at points $A$ and $B$. Let $A D$ be the diameter of $k$ through $A$ and $B P \perp A D, P \in$ $A D$. The line $O D$ meets $B P$ at point $M$. Find the ratio $\frac{B M}{B P}$.

Peter Boyvalenkov
10.1. Let the tangent line to $k$ at $D$ meet the ray $\overrightarrow{O B}$ at point $S$. Then the lines $S D, B P$ and $O A$ are parallel and therefore $\triangle O B M \sim$ $\triangle O S D$ and $\triangle D P M \sim \triangle D A O$. It follows that $\frac{B M}{S D}=\frac{O B}{O S}$ and $\frac{M P}{O A}=\frac{D P}{D A}$, i.e. $B M=$ $\frac{O B \cdot S D}{O S}$ and $M P=\frac{O A \cdot D P}{D A}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-191.jpg?height=373&width=405&top_left_y=121&top_left_x=1176)

Since $O A=O B$ and $\frac{B S}{O S}=\frac{D P}{D A}$ (we use the equality $S D=S B$ ), we obtain $B M=M P$, implying that the desired ratio equals $1: 2$.