[The ratio in which the bisector divides the side.] Law of Cosines

In triangle $ABC$, the bisector of angle $BAC$ intersects side $BC$ at point $M$. It is known that $AB = BC = 2AC$, $AM = 4$. Find the area of triangle $ABC$.
Apply the property of the angle bisector of a triangle and the cosine rule.

By the property of the angle bisector of a triangle,

$$
\frac{B M}{M C}=\frac{A B}{A C}=2
$$

Let $M C=x$. Then

$$
B M=2 x, A B=B C=3 x, A C=\frac{1}{2} A B=\frac{3 x}{2}
$$

Let $P$ be the midpoint of $A C$. From the right triangle $B P C$, we find that

$$
\cos \alpha=\cos \angle B A C=\frac{P C}{B C}=\frac{1}{4}
$$

By the cosine rule,

$$
A M^{2}=A C^{2}+M C^{2}-2 A C \cdot M C \cos \alpha \text {, or } 16=\frac{9 x^{2}}{4}+x^{2}-2 \cdot \frac{3 x}{2} \cdot x \cdot \frac{1}{4} \text {. }
$$

From this equation, we find that $x=\frac{8}{\sqrt{10}}$. Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A C \cdot B C \sin \alpha=\frac{1}{2} \cdot \frac{3 x}{2} \cdot 3 x \cdot \frac{\sqrt{15}}{4}=\frac{18 \sqrt{15}}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-07.jpg?height=460&width=506&top_left_y=1279&top_left_x=773)

## Answer

$\frac{18 \sqrt{15}}{5}$.