[Perpendicular bisector of a segment_([MT)]

[ Inscribed angle subtended by a diameter ]

A circle is circumscribed around triangle $A M B$, and its center is 10 units away from side $A M$.

The extension of side $A M$ beyond vertex $M$ intercepts a segment $C B$ from the tangent to the circle at vertex $B$, which is 29 units long. Find the area of triangle $C M B$, given that the angle $A C B$ is $\operatorname{arctg} \frac{20}{21}$.
Let $O$ be the center of the given circle, $K$ the midpoint of side $AM$, $P$ the projection of point $B$ onto line $AC$, and $F$ the intersection point of ray $BO$ with line $AC$.

Assume that point $O$ lies between points $B$ and $F$. Then

$$
BP = CB \sin \angle C = 29 \cdot \frac{20}{29} = 20
$$

Since $OK = 10$, point $O$ is the midpoint of $BF$, i.e., point $F$ lies on the circle. Therefore, points $A$ and $F$ coincide, and $AB$ is the diameter of the circle. Consequently, points $P$ and $M$ also coincide, and

$$
S_{\triangle CMB} = \frac{1}{2} CM \cdot MB = 10 \sqrt{29^2 - 20^2} = 210.
$$

Similarly, the case where point $F$ lies between points $O$ and $B$ is considered.

## Answer

210.