$\left[\begin{array}{ll}{\left[\begin{array}{l}\text { Law of Cosines }\end{array}\right]} \\ {[\text { Ratio in which the bisector divides the side }]}\end{array}\right]$

Point $O$ is the center of the circle inscribed in isosceles triangle $A B C(A B=B C)$. Line $A O$ intersects segment $B C$ at point $M$. Find the angles and the area of triangle $A B C$, if $A O=3, O M=\frac{27}{11}$.
Apply the property of the bisector of a triangle: the bisector of a triangle divides its side into segments proportional to the other two sides.

## Solution

Since $C O$ is the bisector of triangle $A C M$, then

$$
\frac{C A}{C M}=\frac{A O}{O M}=\frac{11}{9}
$$

Let $A C=11 x$. Then $M C=9 x$. Since $A M$ is the bisector of triangle $B A C$, then $\frac{A B}{A C}=\frac{B M}{M C}$. If $A B=B C$ $=y$,

$$
\frac{y}{11 x}=\frac{y-9 x}{9 x} \text {, or } 9 y=11 y-99 x .
$$

From this, we find that $y=\frac{99 x}{2}$.

If $P$ is the midpoint of $A C$, then

$$
\cos \angle B A C=\cos \angle B C A=\frac{P C}{B C}=\frac{\frac{11 x}{2}}{\frac{99 x}{2}}=\frac{1}{9}
$$

From triangle $A C M$ using the Law of Cosines, we find that

$$
A M^{2}=C A^{2}+C M^{2}-2 C A \cdot C M \cos \angle A C B
$$

or

$$
\left(\frac{60}{11}\right)^{2}=(11 x)^{2}+(9 x)^{2}-2 \cdot 11 x \cdot 9 x \cdot \frac{1}{9}
$$

From this equation, we find that $x^{2}=\frac{20}{121}$. Therefore,

$$
\begin{gathered}
S_{\triangle \mathrm{ABC}}=\frac{1}{2} C A \cdot C B \cdot \sin \angle A C B= \\
=\frac{1}{2} \cdot 11 x \cdot \frac{99 x}{2} \cdot \frac{4 \sqrt{5}}{9}=121 \sqrt{5} x^{2}=20 \sqrt{5}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-39.jpg?height=346&width=546&top_left_y=2514&top_left_x=756)