In triangle $A B C$, it is known that $A B=8, A C=6, \angle B A C=60^{\circ}$. Find the bisector $A M$.
Find the sum of the areas of the triangles into which the bisector divides the given triangle.

## Solution

Since $S_{\triangle \mathrm{ABC}}=S_{\Delta \mathrm{ABM}}+S_{\Delta \mathrm{ACM}}$, then

$$
\frac{1}{2} A B \cdot A C \sin \angle B A C=\frac{1}{2} A B \cdot A M \sin \angle B A M+\frac{1}{2} A M \cdot A C \sin \angle C A M
$$

or

$$
\frac{1}{2} \cdot 8 \cdot 6 \sin 60^{\circ}=\frac{1}{2} \cdot 8 \cdot A M \sin 30^{\circ}+\frac{1}{2} \cdot 6 \cdot A M \sin 30^{\circ}
$$

or $12 \sqrt{3}=\frac{1}{2} \cdot 7 A M$. Therefore, $A M=\frac{24 \sqrt{3}}{7}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-01.jpg?height=415&width=686&top_left_y=1216&top_left_x=686)

## Answer

$\frac{24 \sqrt{3}}{7}$.

## Problem