8,9 In triangle $A B C$, a point $D$ is taken on side $A C$. The circles inscribed in triangles $A B D$ and $B C D$ touch side $A C$ at points $M$ and $N$ respectively. It is known that $A M=3, M D=2, D N=2, N C=4$. Find the sides of triangle $A B C$.
Apply Heron's formula or the cosine theorem.

## Solution

Since $D M=D N$, the circles touch $B D$ at the same point. Let's denote it as $T$. Let $P$ and $Q$ be the points of tangency of the circles with sides $A B$ and $A C, B P=B T=B Q=x$.

By Heron's formula

$$
S_{\triangle \mathrm{ABD}}=\sqrt{(x+5) \cdot x \cdot 2 \cdot 3}, S_{\Delta \mathrm{BCD}}=\sqrt{(x+6) \cdot x \cdot 2 \cdot 4}
$$

On the other hand, $\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{A D}{D C}=\frac{5}{6}$. From the equation

$$
\frac{\sqrt{(x+5) \cdot x \cdot 2 \cdot 3}}{\sqrt{(x+6) \cdot x \cdot 2 \cdot 4}}=\frac{5}{6}
$$

we find that $x=\frac{15}{2}$. Therefore,

$$
A B=3+x=\frac{21}{2}, B C=4+x=\frac{23}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-05.jpg?height=346&width=468&top_left_y=2468&top_left_x=795)

## Answer

## Problem