Let $F' \equiv DP \cap \Gamma$. By symmetry $FF' \parallel CD$. Pascal on $DDF'FEC$ leads to the solution.

Let $M$ be the common midpoint of $AB$ and $CD$ and let $X = CM \cap \Gamma$. We have $C(P,Q;M,D) = C(A,B;M,D) = -1$, so projecting onto $\Gamma$ gives $EXFC$ and $AXBC$ harmonic, and thus $AB, EF, XX$ and $DD$ are concurrent at $G$.

Solution. A humble approach. By symmetry, $DCQP$ is an isosceles trapezoid and hence cyclic, so $D$ is the center of spiral similarity taking $\overline{PQ}$ to $\overline{EF}$, then it also carries $\overline{PE}$ to $\overline{QF}$, implying that $DPEG$ and $DQFG$ are cyclic quadrilaterals. Therefore $$\angle GDE=\angle GPE=\angle DCE$$ thus, $GD$ touches $\Gamma$ at $D$. $\blacksquare$

Generalization: Let $A,B,P,Q,M,N$ be collinear points such that $\{A,B;M,N\}=-1=\{P,Q;M,N\}$. Let $C$ be a point outside line $AB$. Lines $CN,CP,CQ$ cut $\odot ABC$ at $D,E,F$. Lines $EF$ and $AB$ intersect at point $G$. Prove that line $DG$ is tangent to $\odot ABC$. When $N$ is ideal, we have the original problem.

$CDPQ$ - cyclic, $\implies GDPE$ is also cyclic ($D$ is Miquel point of $PQFE$). Since $GP \parallel CD$, by Reim's $GD$ is tangent to $\Gamma$.

Consider the most natural'' projectivity on $\Gamma$ that swaps $E,F$: Projection from $C$ and intersection with line $AB$; reflection about the midpoint of $AB$; and projection again from $C$ onto $\Gamma$. Since se have an involution with corresponding pairs $(E,F),(A,B),(D,D)$, It follows that the lines $AB$, $EF$ and $DD$ (tangent) are concurrent (at $G$).

Posting for Storage. Let $M$ be the midpoint of $BA$ and let $CM \cap \Gamma = T$. $\textbf{Claim:}$ $GT$ is tangent to $\Gamma$. $\textbf{Proof)}$ Let $FM \cap \Gamma = K$, notice by the Butterfly Theorem that $K \in TP$ as well. Now, by Pascal on $TTCEFK$, we get that $G' = TT \cap EF$ lies on $MP$ and therefore on $AB$ meaning that $G'=G$, therefore $GT$ is indeed tangent to $\Gamma$. $\square$ $\textbf{Claim:}$ $GD$ is tangent to $\Gamma$ $\textbf{Proof)}$ Let $P_{\infty}$ denote the point at infinity along $AB$ and $CD$. $$-1 = (A,B;M,P_{\infty}) \stackrel{C}{=}(A,B;T,D)$$ Using the previous $\textbf{Claim}$ which proves that $GT$ is tangent to the circle, we get that $GD$ is also tangent to $\Gamma$ as $G \in AB$. $\blacksquare$

Since $\angle EDA = \angle ECA = \angle PCA = \angle QDB$, $DE$ and $DQ$ are isogonal in $\angle ADB$. So, under the inversion with center $D$ and radius $\sqrt{DA.DB}$, $E,Q$ are swapped and similarly, $F,P$ are swapped. So, $G$ goes to $(DAB) \cap (DPQ) = C$. And since $CD || AB$, $GD$ must be tangent to $(ABD) = \Gamma$, as desired. $\blacksquare$

Note that $\triangle CQB \cong \triangle DPA$, and since $PQ\parallel CD$, we have $CDPQ$ is cyclic. Also note that $\angle DQP = \angle DCP = \angle DCE = \angle DFE \implies DQFG$ is cyclic. Now $\angle GDA = \angle DAB - \angle AGD = \angle DAB - \angle DFC = \angle CAB = \angle DBA$, implying that $DG$ is tangent to $\Gamma$, as desired. $\blacksquare$

Let $E'=DQ\cap \Gamma$. Pascal on $FCDDE'E$ finishes.

Let $M$ be the midpoint of $\overline{AB}$ and $T$ the second intersection of line $\overline{CM}$ with $\Gamma$. Note that $M$ is also the midpoint of $\overline{PQ}$. Let $\infty$ be the point at infinity of $\overline{BC}$. Then, $$\begin{align*} -1&=(AB;M\infty)\stackrel C= (AB;TD)\\ -1&=(PQ;M\infty)\stackrel C= (EF;TD). \end{align*}$$ Thus, the tangents to $\Gamma$ through $T$ and $D$ meet on both lines $\overline{AB}$ and $\overline{EF}$, meaning these tangents meet at $G$, giving the desired result. $\blacksquare$

Let $CM$ meet $\Gamma$ again at $K$. Because $DC \parallel AB$, isogonality implies $DAKB$ is a harmonic quadrilateral. Now, observe $$-1 = (P, Q; M, CD \cap AB) \overset{C}{=} (E, F; K, D)$$ so $DEKF$ is also harmonic. Hence, $DD, KK, AB, EF$ all concur at $AB \cap EF = G$, which implies $G \in DD$, as required. $\blacksquare$

Let $M = \tfrac{A + B}{2} = \tfrac{P + Q}{2}$ and $N = \overline{CM} \cap \Gamma$. Then $$\begin{align*} -1 &= (A, B; M, \infty_{AB}) \overset{C}{=} (A, B; N, D), \\ -1 &= (P, Q; M, \infty_{AB}) \overset{C}{=} (E, F; N, D), \end{align*}$$ which implies that $NN$ and $DD$ must both pass through $G$, where the latter gives the desired conclusion.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.4310156193294326, xmax = 3.916486231576577, ymin = -2.3511733624492717, ymax = 6.134876233881941; /* image dimensions */ /* draw figures */ draw(circle((0,1.865910841994344), 2.3940808821491473), linewidth(0.7)); draw((-2.382364333843369,2.102476797074702)--(2.3823643338433693,2.1024767970747016), linewidth(0.7)); draw((-1.5,0)--(1.5,0), linewidth(0.7)); draw((1.5,0)--(-2.177730037459594,2.8604535279652756), linewidth(0.7)); draw((1.5,0)--(0.9248417364229146,4.074143400590083), linewidth(0.7)); draw((-4.115356167135484,2.102476797074702)--(-1.5,0), linewidth(0.7)); draw((-2.382364333843369,2.102476797074702)--(-4.115356167135484,2.102476797074702), linewidth(0.7)); draw((-4.115356167135484,2.102476797074702)--(0.9248417364229146,4.074143400590083), linewidth(0.7)); draw((-1.5,0)--(-0.9248417364229141,4.074143400590083), linewidth(0.7)); draw((-1.5,0)--(2.1777300374595945,2.860453527965275), linewidth(0.7)); draw((-2.177730037459594,2.8604535279652756)--(-1.5,0), linewidth(0.7)); draw((1.5,0)--(2.1777300374595945,2.860453527965275), linewidth(0.7)); draw((0.9248417364229146,4.074143400590083)--(2.1777300374595945,2.860453527965275), linewidth(0.7)); /* dots and labels */ dot((1.5,0),dotstyle); label("$C$", (1.5611745060097364,-0.27295713477632183), NE * labelscalefactor); dot((2.3823643338433693,2.1024767970747016),dotstyle); label("$B$", (2.5541000373761498,1.9668981328267465), NE * labelscalefactor); dot((-1.5,0),dotstyle); label("$D$", (-2.029334522355703,-0.30986584335773349), NE * labelscalefactor); dot((-2.382364333843369,2.102476797074702),dotstyle); label("$A$", (-2.8183492423000496,1.5745329671523931), NE * labelscalefactor); dot((-1.2031874470492485,2.102476797074702),dotstyle); label("$P$", (-1.082778362396178,2.2209023384312183), NE * labelscalefactor); dot((1.2031874470492487,2.102476797074702),dotstyle); label("$Q$", (1.2302615917384775,1.6460786128616873), NE * labelscalefactor); dot((-2.177730037459594,2.8604535279652756),linewidth(4pt) + dotstyle); label("$E$", (-2.445164556596606,2.9944606009539276), NE * labelscalefactor); dot((0.9248417364229146,4.074143400590083),linewidth(4pt) + dotstyle); label("$F$", (0.9723465746180261,4.172116463301933), NE * labelscalefactor); dot((-4.115356167135484,2.102476797074702),linewidth(4pt) + dotstyle); label("$G$", (-4.223193996485299,2.3132675041055717), NE * labelscalefactor); dot((-0.9248417364229141,4.074143400590083),linewidth(4pt) + dotstyle); label("$F'$", (-0.8749567395520451,4.172116463301933), NE * labelscalefactor); dot((2.1777300374595945,2.860453527965275),linewidth(4pt) + dotstyle); label("$E'$", (2.2192763116828242,2.948278018116751), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Introduce $E' = DQ \cap \Gamma$ and $F' = DP \cap \Gamma$. Then $DPEG$ is cyclic as, $$\angle PDE = \angle FCE' = \frac{1}{2}\left[\widehat{BF} - \widehat{BE'}\right] = \frac{1}{2}\left[ \widehat{BF} - \widehat{AE} \right] = \angle BGF.$$ Then to finish we have, $$\angle F'DG = \angle CEF = \angle DE'F'$$ which is enough.