4. Given a trapezoid $A B C D$ with bases $\overline{A B}=a, \overline{C D}=b, a>b$. Let $M$ and $N$ be the midpoints of the bases $A B$ and $C D$ respectively, and $\overline{M N}=\frac{a-b}{2}$. Calculate the sum of the angles on the larger base.
Solution. We construct the parallelograms $M N C C_{1}$ and $M N D D_{1}$. Then

$$
\begin{array}{r}
\overline{D N}=\overline{N C}=\frac{b}{2}, \overline{A M}=\overline{M B}=\frac{a}{2} \text { and } \\
\overline{M N}=\overline{C C_{1}}=\overline{D_{1}}=\frac{a-b}{2}
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_d711adc0e32ac955c451g-4.jpg?height=257&width=503&top_left_y=525&top_left_x=979)

Furthermore,

$$
\overline{A D_{1}}=\overline{A M}-\overline{M D_{1}}=\frac{a}{2}-\frac{b}{2}=\frac{a-b}{2}=\overline{B M}-\overline{M C_{1}}=\overline{B C_{1}}
$$

Since triangles $A D D_{1}$ and $B C C_{1}$ are isosceles, it follows that $\measuredangle C_{1} D_{1} D=2 \alpha$ and $\measuredangle D_{1} C_{1} C=2 \beta$ (the exterior angle of a triangle is equal to the sum of the two non-adjacent angles). Finally, from $2 \alpha+2 \beta=180^{\circ}$ it follows that $\alpha+\beta=90^{\circ}$.