The circle constructed on the base $A D$ of trapezoid $A B C D$ as its diameter passes through the midpoints of the lateral sides $A B$ and $C D$ of the trapezoid and is tangent to the base $B C$. Find the angles of the trapezoid.
The radius of the circle, drawn to the midpoint of the lateral side, forms an angle of $30^{\circ}$ with the base of the trapezoid.

## Solution

Let $M$ and $N$ be the midpoints of the lateral sides $A B$ and $C D$ of trapezoid $A B C D$, respectively. Then $M N \| A D$.

Let $O$ be the center of the circle, $K$ be the point of tangency with the base $B C$, and $P$ be the point of intersection of the radius $O K$ with the midline $M N$. Then

$$
O P=\frac{1}{2} O K=\frac{1}{2} O N
$$

From the right triangle $P N O$, we find that $\angle P N O=30^{\circ}$. Therefore,

$$
\angle N O D=\angle P N O=30^{\circ}, \angle C D A=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ}
$$

Similarly, we find that $\angle B A D=75^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-48.jpg?height=334&width=640&top_left_y=1748&top_left_x=709)

## Answer

$75^{\circ}, 75^{\circ}, 105^{\circ}, 105^{\circ}$.