Notice that $\measuredangle EDA = \measuredangle CDA = \measuredangle CBA = \measuredangle EAB$ (directed angles), and hence $AB$ is tangent to $\Gamma_1$. Similarly, $AB$ is tangent to $\Gamma_2$. Now notice that since $AP \perp BC$ and $AD \perp BQ$, the contition $PD \perp CQ$ happens if and only if triangles $APD$ and $BCQ$ are similar. This in turn happens if and only if $\frac{AP}{AD} = \frac{BC}{CQ}$, if and only if $AP \cdot CQ = BC \cdot AD$. Now notice that triangles $ADE$ and $BFC$ are similar, and hence $\frac{AD}{AE} = \frac{BF}{BC} \implies AD \cdot BC = AE \cdot BF$. Thus the previous condition turns into $AP \cdot CQ = AE \cdot BF$ which happens if and only if $\frac{AP}{AE} = \frac{BF}{BQ}$. And since we also have $\angle PAE = \angle FBQ = 90^{\circ}$, this in turn happens if and only if $\triangle PAE \sim \triangle FBQ$. Now assume that $\Gamma_1$ and $\Gamma_2$ are tangent at $R$. Straightforward angle manipulation involving the centers of $\Gamma_1$ and $\Gamma_2$ shows that $\angle ARB = 90^{\circ}$ follows from $AB$ being an external common tangent to $\Gamma_1$ and $\Gamma_2$. Thus $A, Q, P, B, R$ are all concyclic on the circle with diameter $AB$. Thus $$\measuredangle ARP = \measuredangle ABP = \measuredangle ABC = \measuredangle ADC = \measuredangle ADE= \measuredangle ARE$$And thus $E, R, P$ are collinear. Analogously $F, R, Q$ are collinear, hence we have $$\measuredangle PEA = \measuredangle REA = \measuredangle RAB = \measuredangle RQB = \measuredangle FQB$$Thus triangles $PEA$ and $FQB$ and similar. Now assume triangles $PEA$ and $FQB$ are similar and let $S$ be the intersection point of $PE$ and $QF$, we shall prove that $\Gamma_1$ and $\Gamma_2$ are tangent at $S$. Due to the similarity we have $$\measuredangle SQB = \measuredangle FQB = \measuredangle PEA = \measuredangle EPC = \measuredangle SPB$$ Hence $S, Q, B, P$ are concyclic and thus $A, Q, P, B, S$ are all concyclic. Now we have $$\measuredangle ASE = \measuredangle ASP = \measuredangle ABP = \measuredangle ABC = \measuredangle ADC = \measuredangle ADE$$ And thus $S$ lies on $\Gamma_1$. Analogously $S$ lies on $\Gamma_2$. Finally: $$\measuredangle PSA = 90^{\circ} = \measuredangle SAB + \measuredangle ABS = \measuredangle SDA +\measuredangle BCS$$ So the tangent line to $\Gamma_1$ at $S$ is also tangent to $\Gamma_2$, and hence $\Gamma_1$ and $\Gamma_2$ are tangent at $S$.