## Task 2 - 260822

Let $k$ be a semicircle over the diameter $A B$. A line intersects $k$ at two points $D$ and $C$, different from $A$ and $B$, and the extension of $A B$ beyond $B$ at a point $E$ such that $C$ lies between $D$ and $E$. Additionally, it is given that

(1) $\overline{B D}=\overline{B E}$ and

(2) $\angle D A C=27^{\circ}$.

Determine the size $\alpha$ of the angle $\angle A C D$!
The triangle $E D B$ is isosceles due to (1), so $\overline{\angle E D B}=\overline{\angle B E D}$.

The angles $\angle A B D$ and $\angle A C D$ are inscribed angles over the same arc $\widehat{A D}$, thus $\overline{\angle A B D}=$ $\overline{\angle A C D}=\alpha$.

The angle $\angle A B D$ is an exterior angle of the triangle $E D B$. Therefore,

$$
\overline{\angle B E D}+\overline{\angle E D B}=\overline{\angle A B D}
$$

and due to (3)

$$
\overline{\angle B E D}=\overline{\angle E D B}=\frac{\alpha}{2}
$$

The angle $\angle B D A$ is a right angle as an inscribed angle over the diameter. In the triangle $A C D$, by the angle sum property and due to (2):

$$
27^{\circ}+\alpha+\frac{\alpha}{2}+90^{\circ}=180^{\circ} \quad \text { so } \quad \alpha=42^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0766.jpg?height=336&width=886&top_left_y=820&top_left_x=585)