[

In triangle $ABC$, the bisector $CD$ is drawn, and the measures of angles $ADC$ and $CDB$ are in the ratio 7:5. Find $AD$, given that $BC=1$ and angle $BAC$ is $30^{\circ}$.

#
property of the bisector of a triangle.

## Solution

Since

$$
\angle A D C=\frac{7}{2} \cdot 180^{\circ}=105^{\circ}
$$

then

$$
\angle A C D=180^{\circ}-30^{\circ}-105^{\circ}=45^{\circ}
$$

Therefore, $\angle A C B=90^{\circ}$, i.e.,

triangle $A B C-$

is a right triangle. Then

$$
\begin{gathered}
A C=B C \operatorname{tg} \angle A B C=\sqrt{3}, A B=2 B C \\
=2 .
\end{gathered}
$$

By the property of the bisector of a triangle

$$
\frac{A D}{D B}=\frac{A C}{B C}=\sqrt{3}
$$

Thus,

$$
\frac{A D}{A B}=\frac{\sqrt{3}}{\sqrt{3}+1}
$$

Therefore,

$$
\begin{aligned}
A D & =\frac{2 \sqrt{3}}{\sqrt{3}+1}=\frac{2 \sqrt{3}(\sqrt{3}-1)}{2} \\
& =\sqrt{3}(\sqrt{3}-1)=3-\sqrt{3}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-12.jpg?height=287&width=560&top_left_y=1743&top_left_x=66)

## Answer

$3-\sqrt{3}$.