On the side $AB$ of the convex quadrilateral $ABCD$, a point $M$ is chosen such that $\angle AMD = \angle ADB$ and $\angle ACM = \angle ABC$. The tripled square of the ratio of the distance from point $A$ to the line $CD$ to the distance from point $C$ to the line $AD$ is $2$, and $CD = 20$. Find the radius of the inscribed circle of triangle $ACD$.
Let $h_{A}$ be the distance from point $A$ to line $C D$, and $h_{C}$ be the distance from point $C$ to line $A D$. According to the condition,

$\frac{h_{A}}{h_{C}}=\sqrt{\frac{2}{3}}$

Since the heights of triangle $A C D$ are inversely proportional to the sides on which they are dropped, then

$A D=C D \cdot \frac{h_{A}}{h_{C}}=20 \sqrt{\frac{2}{3}}$.

Triangles $A M D$ and $A D B$ are similar by two angles. Triangles $A B C$ and $A C M$ are also similar. Therefore, $A M$ :

$A D=A D: A B, A M: M C=A C: A B, \quad$ from which $A D^{2}=A M \cdot A B=A C^{2}$.

Thus, $A D=A C$, meaning triangle $A C D$ is isosceles. Let $p$ be the semiperimeter of this triangle, and $r$ be the radius of the inscribed circle. Then $p=10+20 \sqrt{\frac{2}{3}}, \quad S_{A C D}=\frac{1}{2} C D \sqrt{A D^{2}-\frac{1}{4} C D^{2}}=100 \sqrt{\frac{5}{3}}$. Therefore, $r=\frac{S_{A B C}}{p}=\frac{10 \sqrt{5}}{\sqrt{3}+2 \sqrt{2}}=4 \sqrt{10}-2 \sqrt{15}$.

## Answer

$4 \sqrt{10}-2 \sqrt{15}$