7. On the side $B C$ of triangle $A B C$, a point $K$ is marked.

It is known that $\angle B+\angle C=\angle A K B, \quad A K=5, \quad B K=6, \quad K C=2$.

Find the area of the circle inscribed in triangle $A B C$.
Solution:

$\angle B+\angle C=\angle A K B, \quad \angle A K C=\angle A, \quad \angle K A C=\angle B$,

$\triangle A B C \approx \triangle A K C \Rightarrow$

$\frac{A B}{A K}=\frac{A C}{K C}=\frac{B C}{A C} \Rightarrow \frac{A B}{5}=\frac{A C}{2}=\frac{8}{A C} \Rightarrow$

$A C=4, \quad A B=10$.

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-20.jpg?height=417&width=577&top_left_y=217&top_left_x=1316)

$S_{A B C}=\sqrt{p(p-A B)(p-B C)(p-A C)}, \quad p=\frac{1}{2}(A B+B C+A C)=\frac{1}{2}(10+8+4)=11$,

$S_{A B C}=\sqrt{11(11-10)(11-8)(11-4)}=\sqrt{11 \cdot 3 \cdot 7}=\sqrt{11 \cdot 21} ; \quad r_{s n}=\frac{S_{A B C}}{p}=\frac{\sqrt{21}}{\sqrt{11}} ; \quad S_{\beta n \kappa p}=\frac{21}{11} \pi$.

Answer: $\frac{21}{11} \pi$.