2. $A B C D$ is a trapezoid with bases $A D$ and $B C$, such that a circle with diameter $B C$ can be constructed, passing through the midpoints of the diagonals $A C$ and $B D$, and touching $A D$. Find the angles of the trapezoid.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_719095c7e7bc179570ccg-1.jpg?height=577&width=1190&top_left_y=1750&top_left_x=450)

First method. Let point $M$ be the midpoint of diagonal $A C$, and point $O$ be the center of the circle and the midpoint of base $B C$. In triangle $A B C$, segment $M O$ is the midline, i.e., it is parallel to the lateral side $A B$ of the trapezoid and equal to half of this side. Since the length of $M O$ is equal to the radius $r$ of the circle (point $M$ lies on the circle by condition), then $A B=2 r$. Drop the height $B H$ from point $B$. The leg $B H$ is half the length of the hypotenuse $A B$, therefore, angle $B A D$ is $30^{\circ}$. Then angle $A B C$ is $150^{\circ}$. Similar reasoning for the midpoint of diagonal $B D$ leads to the result $\angle A D C=30^{\circ}, \angle D C B=150^{\circ}$.

Second method. Notice that triangle $B M C$ is a right triangle, as its vertex lies on the circle and its base is the diameter. Therefore, $B M$ is the height in triangle $A B C$, and since this segment is also a median by condition, triangle $A B C$ is isosceles. In the right triangle $A B H$, the hypotenuse is $2 r$, and the leg $B H$ is $r$. We get that angle $B A D$ is $30^{\circ}$. Then angle $A B C$ is $150^{\circ}$. Similar reasoning for the midpoint of diagonal $B D$ leads to the result $\angle A D C=30^{\circ}, \angle D C B=150^{\circ}$.

Answer. $30^{\circ}, 150^{\circ}, 150^{\circ}, 30^{\circ}$