Problem 5. Given trapezoid $ABCD$ with bases $\overline{AB}=16 \text{~cm}, \overline{CD}=8 \text{~cm}$. The sum of angles $ABC$ and $BAD$ is $90^{\circ}$. If $M$ and $N$ are the midpoints of sides $AB$ and $CD$ respectively, calculate the length of segment $MN$.
Solution. Let $P$ and $S$ be points on side $A B$ such that $N P \| A D$ and $N S \| B C$. Triangle $P N S$ is a right triangle, with a right angle at vertex $N$ because

![](https://cdn.mathpix.com/cropped/2024_06_05_bbda9f5c2f069cdaaac4g-12.jpg?height=283&width=758&top_left_y=1685&top_left_x=491)

$$
\measuredangle P S N + \measuredangle S P N = \measuredangle A B C + \measuredangle B A D = 90^{\circ}
$$

$M N$ is a median to the hypotenuse, and therefore

$$
\overline{M N} = \frac{1}{2} \overline{P S} = \frac{1}{2}(16 - 4 - 4) = 4
$$