In trapezoid $A B C D, M$ is the point of intersection of its diagonals. It is known that the lateral side $A B$ is perpendicular to the bases $A D$ and $B C$, and that a circle can be inscribed in the trapezoid. Find the area of triangle $D C M$, if the radius of this circle is $r$.
Let $O$ be the center of the inscribed circle, $K, L$ be the points of tangency with sides $AD$ and $BC$. Recall that $S_{DCM} = S_{ABM}$ (see problem 56764).

Method 1. Let $AD = a$, $BC = b$ be the bases, $h$ be the height of the trapezoid.

$$
\frac{S_{ACD}}{S_{ABC}} = \frac{a}{b}, \frac{S_{MCD}}{S_{ACD}} = \frac{MC}{AC} = \frac{b}{a+b} \Rightarrow S_{DMC} = \frac{ab}{(a+b)^2} S_{ABCD} = \frac{abh}{2(a+b)} \quad \text{(this is true in any trapezoid)}
$$

$CO$ and $DO$ are the angle bisectors of angles $C$ and $D$, so $\angle KDO + \angle LCO = 90^\circ$. Therefore, right triangles $KDO$ and $LMO$ are similar, which means $KD \cdot LC = r^2$. Hence,

$$
\frac{abh}{2(a+b)} = \frac{(r + KD)(r + LC) \cdot 2r}{2(2r + KD + LC)} = \frac{2r^3 + (KD + LC)r^2}{2r + KD + LC} = r^2
$$

Method 2. By a known property of a circumscribed quadrilateral (see problem 57022), the segment $KL$ (connecting opposite points of tangency) passes through point $M$. Point $O$ also lies on this segment. Therefore, $S_{DCM} = S_{ABM} = S_{ABO} = r^2$.

## Answer

$r^2$.