Assume that ray $AB$ dan $DC$ intersects at $X$. Let $R$ be the midpoint of $PQ$, $PQ$ intersects $\Gamma_1$ again at $T$, and $PQ$ intersects $\Gamma_2$ again at $S$. First, we will prove that if $BC$ is parallel to $AD$, then $EF$ bisects segment $PQ$. We will prove that $QT = PS$. Notice that we have $XQ^2 = XC \cdot XD$ and $XP^2 = XB \cdot XA$, so $XP \cdot XQ = \sqrt{XA \cdot XB \cdot XC \cdot XD}$. Since $BC$ is parallel to $AD$, we have $\frac{XB}{XA} = \frac{XC}{XD} \longrightarrow XB \cdot XD = XC \cdot XA$. So, we get $XP \cdot XQ = XA \cdot XC = XB \cdot XD$, and this implies $AP$ is parallel to $QC$ and $DQ$ is parallel to $PB$. Now, we have $\angle{QDC} = \angle{CQB} = \angle{PAB}$, so $DAQP$ is cyclic, so $\triangle{XDQ} \sim \triangle{XAP} \longrightarrow \frac{XD}{XQ} = \frac{XA}{XP}$. Thus, \[\frac{QA}{PC} = \frac{XA}{XP} = \frac{XD}{XQ} = \frac{PD}{QB} \longrightarrow QA \cdot QB = PC \cdot PD\] Since $QA \cdot QB = QT \cdot QP$ and $PC \cdot PD = PQ \cdot PS$, we get $QT = PS$, so $RT = RS$. Hence, $RP \cdot RT = RQ \cdot RS$ and this implies that $R$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$ which is $EF$. Next, we will prove that if $EF$ bisects segmen $PQ$, then $BC$ is parallel to $AD$. So, $R$ lies of $EF$ and this implies that $R$ has equal power to circle $\Gamma_1$ and $\Gamma_2$. Hence, we have $RP \cdot RT = RQ \cdot RS$, so $RT = RS$ and this means $QT = QS$. Notice that $QA \cdot QB = QT \cdot QP = PS \cdot PQ = PC \cdot PD$, thus we get \[(XA - XQ)(XQ - XB) = (XP - XC)(XD - XP)\]\[XA \cdot XQ - XA \cdot XB - XQ^2 + XQ \cdot XB = XP \cdot XD - XP^2 - XC \cdot XD + XC \cdot XP\] Since $XP^2 = XB \cdot XA$ and $XQ^2 = XC \cdot XD$, we get \[XA \cdot XQ + XB \cdot XQ = XD \cdot XP + XC \cdot XP\]\[XQ(XA+XB) = XP(XC+XD)\]\[\frac{XQ}{XP} = \frac{XC+XD}{XA+XB}\] Squaring both sides yields \[\frac{XQ^2}{XP^2} = \frac{XC^2 + 2XC \cdot XD + XD^2}{XA^2 + 2XA \cdot XB + XB^2}\]\[\frac{XC \cdot XD}{XA \cdot XB} = \frac{XC^2 + XD^2 + 2XC \cdot XD}{XA^2 + XB^2 + 2XA \cdot XB}\]\[\frac{XA}{XB} + \frac{XB}{XA} = \frac{XC}{XD} + \frac{XD}{XC}\] \[(\frac{XC}{XD} - \frac{XB}{XA})(\frac{XC}{XD} \cdot \frac{XB}{XA} - 1) = 0\] Thus, $\frac{XC}{XD} = \frac{XB}{XA}$ or $\frac{XC}{XD} \cdot \frac{XB}{XA} = 1$. But since $X$ is the intersection of ray $AB$ and $CD$, we cant have $\frac{XC}{XD} \cdot \frac{XB}{XA} = 1$ because $XC < XD$ and $XB < XA$. So $\frac{XC}{XD} = \frac{XB}{XA}$ and so, $BC$ is parallel to $AD$. Note : The case where $X$ is the intersection of ray $BA$ and $CD$ can be done similarly. I am also looking forward for some non-computation based solution.