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

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-08.jpg?height=694&width=697&top_left_y=1823&top_left_x=585)

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Solution

Since inscribed angles associated with the same arc are equal, we have $\angle D A C=\angle D B C$. Furthermore, knowing that the sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$, it follows that

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

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-09.jpg?height=691&width=714&top_left_y=905&top_left_x=774)

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