A circle is circumscribed around a trapezoid $A B C D$, the side $A B$ is perpendicular to the bases, $M$ is the point of intersection of the diagonals of the trapezoid. The area of triangle $C M D$ is $S$. Find the radius of the circle.
As is known, triangles ВАМ and $D M C$ are equal in area.

## Solution

Let $O$ be the center of the circle; $R$ its radius; $P, F$, and $Q$ the points of tangency with sides $B C, C D$, and $A D$ respectively; $K$ the projection of point $M$ onto side $A B$. Then $C F=C P=B C-B P=B C-R, D F=D Q=A D-A Q=A D-R$.

From the right triangle $C O D$, we get that $R^{2}=O F^{2}=C F \cdot D F=(B C-R)(A D-R)$. Therefore, $R=\frac{A D \cdot B C}{A D+B C}=K M$ (the last equality is proven in the solution to problem $\underline{115592}$).

Since $A B=2 R$, then $S=S_{D M C}=S_{A M B}=\frac{1}{2} A B \cdot K M=R^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-46.jpg?height=445&width=564&top_left_y=2310&top_left_x=747)

## Answer

Problem