Ukrainian National Mathematical Olympiad, 4th Round

Let $E$ be a point on $BC$ such that $AE$ is perpendicular to $CD$. Then $\triangle ACE$ is isosceles, since the bisector of the angle $C$ is also an altitude of the triangle (fig. 23). Hence, $AC = CE$. $\triangle ADE$ is isosceles, since the line $CD$ is perpendicular to $AE$ and divides $AE$ in half (altitude is a median). So, we have:
$$AD = DE. \quad (1)$$
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, $\triangle DEB$ is isosceles, i.e.,
$$BE = DE. \quad (2)$$
From (1) and (2), we have $AD = BE$. Finally, $BC = CE + BE = AC + AD$.