Czech and Slovak Mathematical Olympiad

Since $|BC| = |AP| = |EQ|$, $|BP| = |AQ| = |ED|$ and $|\angle CBP| = |\angle PAQ| = |\angle QED|$, the triangles $PBC$, $QAP$ and $DEQ$ are congruent by the condition SAS.

Hence $|CP| = |PQ| = |QD|$ and also
$$|\angle CPQ| = 180^\circ - |\angle BPC| - |\angle APQ| = 180^\circ - |\angle PQA| - |\angle EQD| = |\angle PQD|.$$
This means that by the condition SAS, the isosceles equilateral triangles $CPQ$ and $DQP$ are also congruent. It follows that their altitudes from $C$ and $D$ to the common opposite side $PQ$ have the same lengths, and hence $CD \parallel PQ$.

Solution 2:

Let $S$ denote the circumcenter of $BAE$. Obviously, $|BA| = |AE|$. Therefore, in the rotation with center $S$ by the oriented angle $BSA$ $B \to A \to E$, and therefore $P \to Q$.

Another consequence of $B \to A \to E$ is the congruence of the four angles $SBA$, $SAB$, $SAE$ and $SEA$. It follows that the angle bisectors of congruent angles $CBA$, $BAE$ and $AED$ are respectively the rays $BS$, $AS$ and $ES$. In our rotation is thus the image of the oriented angle $CBS$ the oriented angle $BAS$, so with respect to $|BC| = |AP|$, $C \to P$ holds. The same is true from the oriented angles $SAE$ and $SED$, the equality $|AQ| = |ED|$ then leads to $Q \to D$. Together we have $C \to P \to Q \to D$, which implies

that the line segments CD and PQ have a common perpendicular bisector—the bisector of CD bisects the angle CSD, and therefore bisects the angle PSQ, and therefore is also the perpendicular bisector of PQ. Because of the common bisector, the lines CD and PQ are parallel.