Let $M$ be the intersection of $AC$ and $BD$ (i.e., the center of $\Gamma_{1}$) and let $N$ be the center of $\Gamma_{2}$. We will first prove that $P, Q$, and $D$ lie on a line. If $P=B$, then $Q=M$ and it is trivial. Otherwise, define $S$ as the intersection of $PQ$ and $BD$. We want to prove that $S=D$. Note that $M, N$, and $P$ lie on a line and that $QN$ is parallel to $DB$. Therefore, by $F$-angles, $\angle PSB = \angle PQN = \angle NPQ$, the last equality due to $|NP| = |NQ|$. With $Z$-angles, we see that $\angle PMB = \angle MNQ$ and by the exterior angle theorem in triangle $PQN$, this is equal to $\angle PQN + \angle NPQ = 2 \angle PSB$. Thus, $\angle PMB = 2 \angle PSB$, from which it follows by the central angle and inscribed angle theorem that $S$ lies on $\Gamma_{1}$. Therefore, $S=D$, which implies that $P, Q$, and $D$ lie on a line.
Since $\angle DPB = \angle DMQ = 90^{\circ}$ and $\angle PDB = \angle QDM$, we have $\triangle DPB \sim \triangle DMQ$. This gives $\frac{|DP|}{|DB|} = \frac{|DM|}{|DQ|}$. Since $|DB| = 2|DM|$, we have $|DP||DQ| = 2|DM|^2$. From the power of a point theorem on $\Gamma_{2}$, it follows that $|DP||DQ| = |DR|^2$, so $|DR| = \sqrt{2}|DM| = |DA|$.