In an isosceles trapezoid with an acute angle $\alpha$ at the base, a circle constructed on the lateral side as a diameter touches the other lateral side.

In what ratio does it divide the larger base of the trapezoid?
Express the specified segments of the larger base in terms of the lateral side of the trapezoid, that is, through the diameter of the circle.

## Solution

Let $O$ be the center of the circle (the midpoint of the lateral side $A B$ of trapezoid $A B C D$), $O P$ be the midline of the trapezoid, and $K$ be the point of intersection of the specified circle with the larger base $A D$. Then $B K \perp A D$ and $K D=O P$. If $M$ is the point of tangency of the circle with the lateral side $C D$, then $\angle M P O=\alpha, K D=O P=O M \operatorname{cosec} \alpha, A K=A B \cos \alpha$. Therefore, ${ }^{A K} /{ }_{K D}=A B \cos \alpha / O M \operatorname{cosec} \alpha=\sin 2 \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-34.jpg?height=386&width=437&top_left_y=670&top_left_x=822)

## Answer

$\sin 2 \alpha: 1$.