Given a convex quadrilateral, if the four angle bisectors of its angles form a new quadrilateral $H I J E$, calculate the sum of the opposite angles $\angle H I J + \angle J E H$.

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-06.jpg?height=594&width=1097&top_left_y=528&top_left_x=385)

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Solution

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-06.jpg?height=591&width=1093&top_left_y=1372&top_left_x=390)

Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have:

$$
\begin{aligned}
\alpha+\beta & =360^{\circ}-\angle I H E-\angle I J E \\
& =360^{\circ}-\angle D H A-\angle C J B \\
& =360^{\circ}-\left(180^{\circ}-\angle A D H-\angle D A H\right)-\left(180^{\circ}-\angle J C B-\angle J B C\right) \\
& =\angle A D H+\angle D A H+\angle J C B+\angle J B C \\
& =\frac{\angle A D C}{2}+\frac{\angle D A B}{2}+\frac{\angle D C B}{2}+\frac{\angle C B A}{2} \\
& =\frac{360^{\circ}}{2} \\
& =180^{\circ}
\end{aligned}
$$