2. Given a trapezoid $A B C D$ such that $\overline{A B}=\overline{A C}=\overline{B D}$. Let $M$ be the midpoint of side $C D$. Determine the angles of the trapezoid if $\measuredangle M B C=\measuredangle C A B$.
Solution. From the condition, it follows that the given trapezoid is isosceles. Let $K$ be the midpoint of $AD$, and let $\measuredangle CAB = \measuredangle MBC = \varphi$. Then

$$
\measuredangle MKA = 180^{\circ} - \measuredangle KAC = 180^{\circ} - \measuredangle MBA.
$$

Thus, quadrilateral $ABMK$ is cyclic.

Furthermore, from the condition, it follows that triangle $\triangle ABD$ is isosceles, from which it follows that $\measuredangle AKB = 90^{\circ}$. Now, due to the cyclicity of $ABMK$, it follows that $\measuredangle AMB = \measuredangle AKB = 90^{\circ}$, i.e., triangle $\triangle AMB$ is an isosceles right triangle. Let $M_1$ be the foot of the altitude dropped from $M$. Then

$$
\overline{MM_1} = \overline{AM_1} = \frac{\overline{AB}}{2} = \frac{\overline{AC}}{2},
$$

so we get that $\varphi = 30^{\circ}$. Now easily

$$
\measuredangle ABC = 30^{\circ} + 45^{\circ} = 75^{\circ} \text{ and } \measuredangle ADC = 105^{\circ}.
$$