In triangle $A B C$, side $A B$ is equal to 21, the bisector $B D$ is $8 \sqrt{7}$, and $D C=8$. Find the perimeter of triangle $A B C$.
Apply the property of the bisector of a triangle and the sine and cosine theorems.

## Solution

Let $A D=x, B C=y, \angle A B D=\angle C B D=\alpha$.

By the property of the bisector of a triangle, $\frac{A B}{B C}=\frac{A D}{D C}$, or $\frac{21}{y}=\frac{x}{8}$, from which we find that $y=\frac{21 \cdot 8}{x}$. By the sine theorem for triangles $A B C$ and $D B C$, we have:

$$
\frac{A C}{\sin \angle A B C}=\frac{A B}{\sin \angle C}, \frac{D C}{\sin \angle C B D}=\frac{B D}{\sin \angle C}
$$

or

$$
\frac{x+8}{\sin 2 \alpha}=\frac{21}{\sin \angle C}, \frac{8}{\sin \alpha}=\frac{8 \sqrt{7}}{\sin \angle C}
$$

Dividing the first equation by the second and applying the formula $\sin 2 \alpha=2 \sin \alpha \cos \alpha$, we get

$$
\frac{x+8}{2 \cos \alpha}=\frac{21}{\sqrt{7}}
$$

from which $\cos \alpha=\frac{x+8}{6 \sqrt{7}}$.

By the cosine theorem from triangle $A B D$, we find that

$$
x^{2}=21^{2}+(8 \sqrt{7})^{2}-2 \cdot 21 \cdot 8 \sqrt{7} \cdot \frac{x+8}{6 \sqrt{7}}
$$

or

$$
x^{2}+56 x-441=0
$$

from which $x=7$. Then $y=\frac{21 \cdot 8}{x}=24$. Therefore, the perimeter of triangle $A B C$ is

$$
A B+A C+B C=A B+(A D+C D)+B C=21+7+8+24=60
$$

## Answer

60.