## Task A-3.4.

In a convex quadrilateral $A B C D$, $\varangle B A D=50^{\circ}, \varangle A D B=80^{\circ} \text{ and } \varangle A C B=40^{\circ}$.

If $\varangle D B C=30^{\circ}+\varangle B D C$, calculate $\varangle B D C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_659ee8c29a87e0ecc7d7g-16.jpg?height=534&width=825&top_left_y=515&top_left_x=547)

Since

$$
\varangle A B D=180^{\circ}-80^{\circ}-50^{\circ}=50^{\circ}=\varangle B A D, \quad 1 \text { point }
$$

triangle $A B D$ is isosceles and $|A D|=|B D|$ holds.

Since $|A D|=|B D|$ and $\varangle A D B=2 \cdot \varangle A C B$, point $D$ is the center of the circumscribed circle of triangle $A B C$.

Let $\varangle B D C=x$. Since angle $B A C$ is an inscribed angle over the chord $\overline{B C}$, and the corresponding central angle is angle $B D C$, we have $\varangle B A C=\frac{x}{2}$.

Since $\varangle C B D=x+30^{\circ}$, from triangle $A B C$ it follows that

$$
\begin{aligned}
180^{\circ} & =\frac{x}{2}+50^{\circ}+x+30^{\circ}+40^{\circ} \\
\text { i.e. } x & =40^{\circ}
\end{aligned}
$$