3. In isosceles $\triangle A B C$ the perpendicular bisector of side $A C$ and the angle bisector of $\measuredangle B A C$ intersect at point $D$ which lies on side $B C$. Determine the measure of $\Varangle C D A$.
Solution. From the condition of the problem, it follows that $\measuredangle B A C=\measuredangle A B C=\beta$. Furthermore, $A D$ is the angle bisector of $\measuredangle B A C$, so $\measuredangle B A D=\measuredangle D A C=\frac{\beta}{2}$. However, point $D$ lies on the perpendicular bisector of side $A C$, so $\overline{A D}=\overline{C D}$, i.e., triangle $C A D$ is isosceles with base $AC$. Therefore, $\measuredangle C A D=\measuredangle D A C=\frac{\beta}{2}$. According to this, for the sum of the angles of $\triangle A B C$ we get

![](https://cdn.mathpix.com/cropped/2024_06_05_b088eb3db01806ad63fdg-04.jpg?height=490&width=353&top_left_y=1243&top_left_x=1107)

$$
\beta+\beta+\frac{\beta}{2}=180^{\circ} \text {, i.e., } \beta=72^{\circ} \text {. }
$$

Finally, from $\triangle C A D$ we find

$$
\measuredangle C D A=180^{\circ}-2 \cdot \frac{\beta}{2}=180^{\circ}-\beta=180^{\circ}-72^{\circ}=108^{\circ}
$$