$[\mathrm{O}$ The Law of Cosines ] [ The ratio in which the bisector divides the side]

In triangle $A B C$, the bisector of angle $A B C$ intersects side $A C$ at point $K$. It is known that $B C=2, K C=1, B K=$ $\frac{3 \sqrt{2}}{2}$. Find the area of triangle $A B C$.
Find the cosine of angle $C$ and use the property of the angle bisector of a triangle.

## Solution

By the cosine theorem from triangle $B K C$, we find that

$$
\cos \angle C=\frac{B C^{2}+K C^{2}-B K^{2}}{2 B C \cdot K C}=\frac{1}{8}
$$

By the property of the angle bisector of a triangle,

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

Let $A K=x$. Then $A B=2 x$, and by the cosine theorem from triangle $A B C$, we find that

$$
A B^{2}=A C^{2}+B C^{2}-2 A C \cdot B C \cos \angle C, \text { or } 4 x^{2}=(x+1)^{2}+4-2 \cdot 2 \cdot(x+1) \cdot \frac{1}{8}
$$

From this equation, we find that $x=\frac{3}{2}$. Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} B C \cdot A C \sin \angle C=\frac{15 \sqrt{7}}{16}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-16.jpg?height=383&width=638&top_left_y=0&top_left_x=710)

## Answer

$\frac{15 \sqrt{7}}{16}$.