## Task A-1.3.

Let $A B C$ be a triangle in which $\varangle C A B=20^{\circ}$ and let $D$ be the midpoint of side $\overline{A B}$.

If $\varangle C D B=40^{\circ}$, determine $\varangle A B C$.
## Solution.

Since $\varangle B D C=40^{\circ}$, it follows that $\varangle C D A=140^{\circ}$ and

$$
\varangle A C D=180^{\circ}-\varangle D A C-\varangle C D A=20^{\circ}
$$

Therefore, triangle $A D C$ is isosceles, so $|C D|=|A D|=|B D|$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-02.jpg?height=365&width=791&top_left_y=1922&top_left_x=564)

Finally, this means that triangle $B D C$ is isosceles, so

$$
\varangle C B D=\varangle D C B=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ} .
$$