7. In triangle $A B C$, angles $A$ and $B$ are $45^{\circ}$ and $30^{\circ}$ respectively, and $C M$ is the median. The incircles of triangles $A C M$ and $B C M$ touch segment $C M$ at points $D$ and $E$. Find the radius of the circumcircle of triangle $A B C$ if the length of segment $D E$ is $4(\sqrt{2}-1)$.
# Solution:

By the property of tangents to a circle, we have:

$$
A G=A K=x, C G=C D=y, C E=C F=z, B F=B H=u, D M=\frac{A B}{2}-A K=\frac{A B}{2}-x
$$

$M E=\frac{A B}{2}-B H=\frac{A B}{2}-u$,

Then $D E=z-y, D E=D M-M E=u-x$. Therefore, $2 D E=z-y+u-x=C B-$ AC. Let $C B=a, A C=b$. Then $a-b=8(\sqrt{2}-1)$. By the Law of Sines, we have

$\frac{a}{\sin 45^{\circ}}=\frac{b}{\sin 30^{\circ}}$, or $\sqrt{2} a=2 b$,

i.e., $a=\sqrt{2} b$. Thus,

$b=8, a=8 \sqrt{2}$.

The radius of the circumcircle of triangle

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-14.jpg?height=437&width=1013&top_left_y=1483&top_left_x=927)
$A B C$ is found using the formula

$R=\frac{b}{2 \sin 30^{\circ}}=b=8$.

## Answer: 8.