Macedonian Junior Mathematical Olympiad

By the conditions of the task it follows that the trapezoid is isosceles. Let $K$ be the midpoint of $AD$, and let $\angle CAB=\angle MBC=\phi$. Then $$\angle MKA=180^{\circ }-\angle KAC=180^{\circ }-\angle MBA.$$ Therefore the quadrilateral $ABMK$ is inscribed. Then, by the conditions we have that $△ABD$ is isosceles, from where we get $\angle AKB=90^{\circ }$. Now, because of the fact that $ABMK$ is inscribed, we have $\angle AMB=\angle AKB=90^{\circ }$ i.e. we get that the triangle $△AMB$ is a right isosceles triangle. Let $M_1$ be the foot of the altitude from $M$. Then $$\overline{M{M}_1}=\overline{A{M}_1}=\frac{\overline{AB}}{2}=\frac{\overline{AC}}{2},$$ so we get that $\phi =30^{\circ }$. Now we easily get $\angle ABC=30^{\circ }+45^{\circ }=75^{\circ }$ and $\angle ADC=105^{\circ }$.