7. In triangle $A B C$, angles $A$ and $B$ are $45^{\circ}$ and $30^{\circ}$ respectively, and $C M$ is a median. The incircles of triangles $A C M$ and $B C M$ touch segment $C M$ at points $D$ and $E$. Find the area 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:

$$
A G=A K=x, C G=C D=y
$$

$C E=C F=z, B F=B H=u$,

$D M=\frac{A B}{2}-A K=\frac{A B}{2}-x$,

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

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-08.jpg?height=428&width=1008&top_left_y=748&top_left_x=981)

Then $D E=z-y, D E=D M-M E=u-x$. Therefore, $2 D E=z-y+u-x=C B-A C$. 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 area of triangle $A B C$ is found using the formula $S=\frac{a b \sin 105^{\circ}}{2}$.

$$
S=\frac{a b \sin \left(45^{\circ}+30^{\circ}\right)}{2}=\frac{a b\left(\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}\right)}{2}=\frac{a b \sqrt{2}(\sqrt{3}+1)}{8}=16(\sqrt{3}+1) .
$$

Answer: $16(\sqrt{3}+1)$.