[ Sum of angles in a triangle. Theorem about the exterior angle.]

$A B C$ is an isosceles triangle with base $A C, C D$ is the bisector of angle $C, \angle A D C=150^{\circ}$. Find $\angle B$.

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Apply the theorem about the sum of the angles of a triangle.

## Solution

Let $\angle A=\angle C=2 \alpha$. Then $\angle D C A=\angle B C D=\alpha$. By the condition $2 \alpha+\alpha=180^{\circ}-150^{\circ}$, from which $\alpha=10^{\circ}$. Therefore, $\angle B=180^{\circ}-4 \alpha=140^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d7c3832c1c4b0d15493ag-02.jpg?height=306&width=918&top_left_y=2157&top_left_x=570)

Answer

$140^{\circ}$.

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