62nd Ukrainian National Mathematical Olympiad

Let the rays $AB$ and $DE$ intersect at the point $O$ (fig. 16). Then $BCDO$ is a parallelogram, and $\Delta AOE$ is isosceles, since $\angle OAE=\angle AEO$. So, $$AB+BC=(OB-OA)+BC=CD-OE+OD=CD+DE.$$