Problem 2. Given $\triangle A B C$ such that $\measuredangle A B C - \measuredangle A C B = 30^{\circ}$. On side $A C$, point $D$ is taken such that $\overline{A B} = \overline{A D}$. Determine the measure of angle $C B D$.
Solution. We have

$$
\begin{gathered}
\measuredangle A B C=30^{\circ}+\measuredangle A C B, \measuredangle C A B=180^{\circ}-(\measuredangle A C B+\measuredangle A B C), \\
\measuredangle C A B=180^{\circ}-\left(30^{\circ}+2 \measuredangle A C B\right)=150^{\circ}-2 \measuredangle A C B .
\end{gathered}
$$

Since $\overline{A B}=\overline{A D}$, triangle $\triangle A D B$ is isosceles, so

$$
\begin{aligned}
\measuredangle A D B & =\measuredangle A B D=\frac{180^{\circ}-\measuredangle \angle A B}{2} \\
& =\frac{30^{\circ}+2 \measuredangle A C B}{2}=15^{\circ}+\measuredangle A C B
\end{aligned}
$$

According to this,

![](https://cdn.mathpix.com/cropped/2024_06_05_b353751bdd87cc544500g-09.jpg?height=209&width=574&top_left_y=1496&top_left_x=886)

$$
\begin{aligned}
& \Varangle A B C=\measuredangle A B D+\measuredangle D B C \\
& 15^{\circ}+\measuredangle A C B+\measuredangle C B D=30^{\circ}+\measuredangle A C B \\
& \measuredangle C B D=15^{\circ}
\end{aligned}
$$