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

In triangle $ABC$, the bisectors $BL$ and $AE$ of angles $ABC$ and $BAC$ respectively are drawn, intersecting at point $O$. It is known that $AB=BL$, the perimeter of triangle $ABC$ is $28$, and $BO=2OL$. Find $AB$.
Use the property of the bisector of a triangle and the cosine theorem.

## Solution

Let $O L=2 a, B O=4 a$. Then $A B=B L=6 a$. Since $A O$ is the bisector of triangle $A B L$, then $\frac{A B}{A L}=\frac{B O}{O L}=2$, so $A L=\frac{1}{2} A B=3 a$.

Let $C L=b$. Since $B L$ is the bisector of triangle $A B C$, then $\frac{B C}{C L}=\frac{A B}{A L}=2$, so $B C=2 C L=2 b$.

The height $B K$ of the isosceles triangle $A B L$ bisects the base $A L$, so $\cos \angle B A C=\frac{A K}{A B}=\frac{1}{4}$. By the cosine theorem,

$$
\begin{gathered}
B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \angle B A C, \text { or } \\
4 b^{2}=36 a^{2}+(3 a+b)^{2}-3 a(3 a+b) .
\end{gathered}
$$

After obvious transformations, we get the equation

$$
b^{2}-a b-12 a^{2}=0
$$

from which we find that $b=4 a$. Then the perimeter of triangle $A B C$ is

$$
A B+B C+A C=6 a+8 a+7 a=21 a=28
$$

from which $a=\frac{4}{3}$. Therefore, $A B=6 a=8$.

## Answer

8.