## Task A-4.3.

In a convex quadrilateral $A B C D$, the following holds:

$$
\varangle B A C=48^{\circ}, \quad \varangle C A D=66^{\circ}, \quad \varangle C B D=\varangle D B A
$$

Determine the angle $\varangle B D C$.
## First Solution.

Let $\varphi=\varangle C B D=\varangle D B A$. From triangle $A B D$ we have:

$$
\begin{aligned}
\varangle A D B & =180^{\circ}-\varangle B A D-\varangle D B A \\
& =180^{\circ}-(\varangle B A C+\varangle C A D)-\varangle D B A \\
& =180^{\circ}-\left(48^{\circ}+66^{\circ}\right)-\varphi \\
& =66^{\circ}-\varphi .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_12fcd97ef28da03e8b64g-18.jpg?height=885&width=897&top_left_y=1708&top_left_x=588)

If we mark a point $E$ on the diagonal $\overline{B D}$ such that $\varangle E A D=\varangle A D E=\varangle A D B=66^{\circ}-\varphi$, we conclude that triangle $E D A$ is isosceles with base $\overline{D A}$ and that $\varangle C A E=\varphi=\varangle D B A$. Also, $\varangle C A E=\varphi=\varangle C B D=\varangle C B E$, so quadrilateral $A B C E$ is cyclic. Thus, $\varangle E C A=\varangle E B A=\varangle D B A=\varphi$, and triangle $E A C$ is isosceles with base $\overline{A C}$.

Since $|E A|=|E C|=|E D|$, point $E$ is the center of the circumcircle of triangle $A C D$ and by the theorem of the inscribed and central angles, $\varangle C E D=2 \varangle C A D=132^{\circ}$.

Finally, $\varangle B D C=\varangle E D C=\varangle D C E=\frac{1}{2}\left(180^{\circ}-\varangle C E D\right)=24^{\circ}$.