We assume that $D,E$ lie on the side-segments $AB,AC.$ Denote $\omega_1 \equiv \odot(APD)$ and $\omega_2 \equiv \odot(AQE).$ Because of $\angle SQE=\angle SAE=\angle SAB,$ it follows that $B,S,Q$ are collinear. Likewise, $C,R,P$ are collinear. Thus, $SE \parallel RD \parallel BC$ $\Longrightarrow$ arcs $ES$ and $RD$ of $\omega_2$ and $\omega_1$ are homologous $\Longrightarrow$ $ER,DS$ intersect at the exsimilicenter of $\omega_1,\omega_2.$ On the other hand, since the tangent line $\tau$ to $\Gamma$ through $A,$ $AA'$ and $BC$ bound an isosceles triangle, then $\tau$ cuts $\omega_2$ again at $F,$ such that $AFES$ is an isosceles trapezoid, i.e. $EF \parallel RA$ $\Longrightarrow$ $F,A$ are homologous points under the positive homothety that takes $\omega_1$ into $\omega_2.$ Hence, $DS,ER,\tau$ concur at the exsimilicenter of $\omega_1,\omega_2.$

Quote: Arcs $ES$ and $RD$ of $\omega_2$ and $\omega_1$ are homologous $\Longrightarrow$ $ER,DS$ intersect at the exsimilicenter of $\omega_1,\omega_2.$ Luis, what is homologous and why $ ER,DS $ intersect at the exsimilicenter?

What ia homogolous??? Is there another sinplere and more elemntary solution????

My solution: Let $ T $ be the intersection of $ DS $ and $ RE $ Let $ X $ be the intersection of $ TA $ and $ (AQE) $ Easy to see $ DR \parallel SE \parallel BC $ (Reim theorem). Since arc $ DR $ = arc $ SE $ , so $ DR $ and $ SE $ are corresponding chord in $ (APD) $ and $ (AQE) $ , hence we get $ T $ is the exsimilicenter of $ (APD) \sim (AQE) $ . Since $ \triangle ADR $ and $ \triangle XSE $ are homothetic with center $ T $ , so $ \angle TAB =\angle TXS =180^{\circ}-\angle AES =180^{\circ}-\angle ACB $ . ie. $ TA $ is tangent to $ (ABC) $ Q.E.D

Note that by Reim's theorem we have $DR \parallel SE \parallel BC$. Let $PR\cap\Gamma=X$, Then note that by Reim's theorem again we have $DR \parallel BX \implies X=C$ and $P,R,C$ collinear, similarly $Q,S,B$. Let the tangent at A intersect the line $PQ$ at $T$. Then by pascal on $AABQPA'$ we have $T, D,S$ collinear, as are $T,R,E$ by pascal on $AACPQA'$, so the three lines are concurrent.

Through elementary angle-chasing, we can see that $\measuredangle BQA' = \measuredangle BAA' = \measuredangle SAE = \measuredangle SQE = \measuredangle SQA'$. Thus, points $B,S,Q$ must be collinear ($\measuredangle BQA' = \measuredangle SQA'$ implies that both $B$ and $S$ lie on the same line passing through $Q$). Similarly, we can see that points $P,R,C$ are collinear. Thus, $\measuredangle BCA = \measuredangle BQA = \measuredangle SQA = \measuredangle SEA$, implying that $BC || SE$. Similarly, we can use the collinearity described above between points $P,R,C$ to show that $DR || BC$. This means that $DR || SE$, implying that the point $DS \cap RE$ is the center of a homothety taking $DR$ to $SE$. Let $Y$ be the second intersection of the tangent to $A$ with $\odot SEQA$. Now, we can again use angle chasing to show that $\measuredangle BAP = \measuredangle BCA = \measuredangle SEA = \measuredangle SYA$, and therefore $DA || SY$. This means that the homothety described above not only takes $DR$ to $SE$, but in fact $\triangle DAR$ to $\triangle SYE$!! Another way of saying this is that the center of this homothety lies at $AY \cap DS \cap ER$, which implies that these three lines are concurrent. Q.E.D

Here is my solution: Let's say $D,E$ lie on the side-segments $AB,AC$ respectively and denote $\omega_1 \equiv \odot(APD)$ and $\omega_2 \equiv \odot(AQE).$ By simple angle chasing we can observe that $B,S,Q$ are collinear likewise, $C,R,P$ are collinear. Thus, $SE \parallel RD \parallel BC$ . Let's define $K$ as $SK \parallel DA$ and $EK \parallel AR$ . Since all $3$ sides are parallel to each other triangle $ADR$ and triangle $KSE$ are similar $\Longrightarrow$ $\angle SKE=\angle DAR=\angle SAE$ it follows that $K$ lies on $\omega_1$ . Now we can say $\angle EAK=\angle KSE=\angle ABC$ $\Longrightarrow$ $AK$ is tangent to $\Gamma$ at $A$ . We are ready to finish by applying desargues (or homothety) to triangle $ADR$ and $KSE$ . ( $AK$, $DS$ and $RE$ needs to concur)

Let $A'D\cap BC=D_1,A'E\cap BC=E_1,AA'\cap BC=K,RE\cap SD=T$. Let $W\in (ABC)$ with $AW\parallel BC$. Since $A'D.A'P=A'R.A'A$ and $A'D_1.A'P=A'K.A'A$ we get $DR\parallel BC$. Similarily $ES\parallel BC$. Apply DDIT on quadrilateral $RDSE$ to get $(\overline{AT},\overline{A(BC)_{\infty}}),(\overline{AD},\overline{AE}),(\overline{AA'},\overline{AA'})$ is an involution. Project this onto $(ABC)$. We see that $A'A',BC,(AT\cap (ABC))W$ are concurrent. Since $A'A'\cap BC=BC_{\infty}$, we observe that $AT\cap (ABC)=A$ so $AT$ is tangent to $(ABC)$ as desired.$\blacksquare$