59th Ukrainian National Mathematical Olympiad

On the ray $BA$, we put down a segment $BE=BC$. If point $E$ belongs to $AB$ (fig. 19), then $△BCD=△BED$ due to two pairs of equal sides and the angle between them. Then, $AD=CD=ED$, which yields that $△ADE$ is isosceles. There, $HD$ is the height and, hence, the median. Therefore,

$$AH=HE=HB-BE=HB-BC\Rightarrow BC=BH-HA.$$

If point $A$ belongs to the segment $BE$ (fig. 20), then, analogously, $△BCD=△BED$ and

$$AH=HE=BE-HB=BC-HB\Rightarrow BC=BH+HA.$$