Junior Balkan Mathematical Olympiad

$$AB+CD=(AP+PB)+(CR+RD)=AP+(BP+CR+DR)$$ $$BC+DE=(BQ+QC)+(DS+SE)=ES+(BQ+CQ+DS).$$ Using the fact that tangents from a point to a circle are of equal length, one gets $AP=ES$. Denoting by $O$ and $r$ the center, respectively radius of $k$, one gets that right-angled triangles $OPA$, $OSE$ are congruent, since $AP=ES$, $OP=OS=r$, and $OA=\sqrt{O{P}^2+A{P}^2}=\sqrt{O{S}^2+E{S}^2}=OE$. Therefore the corresponding altitudes of these triangles, from $P$, respectively $S$, are of equal length, hence $PS$ is parallel to $AE$.