Clearly $R$ belongs to the bisector line of $\angle CBA$. Let's suppose $A,P,Q,C$ concyclic in a circle $\Gamma_3$. Then, from the existence of the radial center of crcles $\Gamma_1$, $\Gamma_2$ and $\Gamma_3$, one gets that $PA$, $QC$ and $BR$ concur, which is the same as saying that $DA$ and $EC$ intersect at the bisector line of angle $\angle CBA$. But, from the definition of $D$ and $E$, we get that $ACED$ is an isosceles trapezoid, and so $DA$ and $EC$ intersect in the perpendicular bisector line of segment $AC$. Hence the perpendicular bisector line of $AC$ coincides with the bisector line of $\angle CBA$, which implies $AB=BC$. Now, let's suppose $AB=BC$. Then, points $D,E$ and $P,Q$ are symmetrical with respect to the perpendicular bisector line of segment $AC$. Hence $APQC$ is an isosceles trapezoid, always cyclic.

Since $AC \parallel DE$, $ACED$ is isosceles trapezoid, then $\angle DAC = \angle ACE$. Notice that $\angle DPR = \angle APR = \angle ABR = \angle RBC = \angle RQC = \angle RQE$ and $\angle PDR = \angle PDE = \angle DAC = \angle DEQ = \angle REQ$ thus $\triangle DPR \cong \triangle EQR.$ By Reim's Theorem $APQC$ is cyclic $\Leftrightarrow DEQP$ is cyclic. We have $APQC$ is cyclic $\Leftrightarrow DEQP$ is isosceles trapezoid $\Leftrightarrow \angle DPQ = \angle EQP$ $\Leftrightarrow \angle RPQ = \angle RQP$ (since $\angle DPR = \angle RQE$) $\Leftrightarrow \triangle RPQ$ is isosceles. Now, $\angle PAB = \angle BRP = \angle BRQ = \angle BCQ$. Therefore $APQC$ is cyclic $\Leftrightarrow RPQ$ is isosceles $\Leftrightarrow \angle BAC = \angle ACB$ (since $\angle BAC = \angle ACE$) $\Leftrightarrow AB = BC$ as desired.

Attachments: