## Task A-2.4.

Points $A, B, C, D, E$ lie in that order on a circle whose diameter is $\overline{A E}$.

Determine $\varangle A B C + \varangle C D E$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_0bfcef02cf5258b8afa5g-11.jpg?height=562&width=609&top_left_y=507&top_left_x=655)

Quadrilateral $ABCE$ is cyclic, so $\varangle ABC = 180^{\circ} - \varangle AEC$. 2 points

Similarly, quadrilateral $ACDE$ is cyclic, so $\varangle CDE = 180^{\circ} - \varangle CAE. \quad 1$ point

By Thales' theorem, triangle $ACE$ is right-angled, so $\varangle CAE + \varangle AEC = 90^{\circ}. \quad 1$ point

Now, by adding these, we get $\varangle ABC + \varangle CDE = 360^{\circ} - 90^{\circ} = 270^{\circ}$.