3. In triangle $ABC (\overline{BC} > \overline{AC})$, the angles $\alpha = \measuredangle BAC$ and $\beta = \measuredangle ABC$ differ by $30^{\circ}$. If point $D$ is chosen on side $BC$ such that $\overline{AC} = \overline{CD}$, calculate $\measuredangle DAB$.
Solution. Let $\measuredangle D A B=x$. Triangle $A C D$ is isosceles, and it holds that $\measuredangle D A C=\measuredangle A D C=\alpha-x$.

Then from triangle $A C D$ we have:

$$
\begin{aligned}
\alpha-x+\alpha-x+\gamma & =2 \alpha-2 x+180^{\circ}-\alpha-\beta \\
& =180^{\circ}
\end{aligned}
$$

which means $\alpha-\beta=2 x$, from which it is clear that

![](https://cdn.mathpix.com/cropped/2024_06_05_8cdbd8422ee96a9c77ddg-2.jpg?height=353&width=469&top_left_y=434&top_left_x=991)
$x=15^{\circ}$.