## Task A-2.4.

The vertices of the pentagon $A B C D E$ lie on the same circle. If $\varangle C A D=50^{\circ}$, determine

$$
\varangle A B C+\varangle A E D \text {. }
$$
## First solution.

Angles $\varangle C A D$ and $\varangle C E D$ are inscribed angles subtending the same arc, so $\varangle C E D = \varangle C A D = 50^{\circ}$.

Quadrilateral $A B C E$ is cyclic, so $\varangle A B C + \varangle A E C = 180^{\circ}$.

Therefore, we have

$$
\begin{aligned}
\varangle A B C + \varangle A E D & = \varangle A B C + (\varangle A E C + \varangle C E D) \\
& = (\varangle A B C + \varangle A E C) + \varangle C E D \\
& = 180^{\circ} + 50^{\circ} = 230^{\circ} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_acf83088931d085246bdg-10.jpg?height=803&width=766&top_left_y=1106&top_left_x=568)