5. Inside an isosceles triangle $A B C$ with the angle at the vertex $\measuredangle A C B=100^{\circ}$, there is a point $D$ such that $\measuredangle B A D=20^{\circ}$ and $\measuredangle A B D=30^{\circ}$. Find $\measuredangle B C D$!
Solution. Let $O$ be the intersection of line $B D$ with the altitude $C H$. In this way, we obtain the isosceles $\triangle A B O$, with base angles of $30^{\circ}$. From this, we conclude that $\measuredangle O A D=10^{\circ}$, and since

![](https://cdn.mathpix.com/cropped/2024_06_05_162556eb92784f75b3beg-4.jpg?height=468&width=820&top_left_y=1381&top_left_x=508)

$$
\measuredangle B A C=\measuredangle A B C=\frac{180^{\circ}-100^{\circ}}{2}=40^{\circ} \Rightarrow \measuredangle O A C=40^{\circ}-30^{\circ}=10^{\circ}
$$

The angle $\measuredangle O D A=50^{\circ}$, as an exterior angle for $\triangle A B D$, and also $\measuredangle A C O=\frac{1}{2} \cdot 100^{\circ}=50^{\circ}$. It follows that $\triangle A O C \cong \triangle A O D$ (ASA), from which we get $\overline{A C}=\overline{A D}$, meaning $\triangle A C D$ is isosceles, with base $C D$ and base angles of $80^{\circ}$ (since the vertex angle is $20^{\circ}$). Then for the desired angle we get: $\measuredangle B C D=\measuredangle A C B-\measuredangle A C D=100^{\circ}-80^{\circ}=20^{\circ}$.

## VIII Section