51st Ukrainian National Mathematical Olympiad, 4th Round

Let $E$ be a point on $BC$ such that $AE$ is perpendicular to $CD$. Then $△ACE$ is isosceles, since the bisector of the angle $C$ is also an altitude of the triangle (fig. 23). Hence, $AC=CE$. $△ADE$ is isosceles, since the line $CD$ is perpendicular to $AE$ and divides $AE$ in half (altitude is a median). So, we have: $$AD=DE.(1)$$ Let $\angle B=\alpha$, $\angle A=2\alpha$, $\angle C=180^{\circ }-3\alpha$. Then $\angle ADC=\alpha +90^{\circ }-\frac{3\alpha }{2}=90^{\circ }-\frac{\alpha }{2}$, so $\angle ADE=180^{\circ }-\alpha$, $\angle EDB=\alpha =\angle B$, hence, $△DEB$ is isosceles, i.e., $$BE=DE.(2)$$ From (1) and (2), we have $AD=BE$. Finally, $BC=CE+BE=AC+AD$.