[asy][asy] import graph; size(11.3cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -2.29, xmax = 5.24, ymin = -1.65, ymax = 2.88; /* image dimensions */ pen uququq = rgb(0.25,0.25,0.25); pen xdxdff = rgb(0.49,0.49,1); pair P = (0,0), A = (0.76,0.66), B = (0.88,-0.48), Q = (1.21,0.13), C = (1.26,0.82), D = (1.16,-0.56), F = (0.43,-0.9), T = (0.98,-0.29), G = (-1,0.08), H = (0.83,-0.06), I = (1,-0.08), J = (-0.81,2.46); /* draw figures */ draw(circle(P, 1)); draw(circle(Q, 0.69)); draw(D--F); draw(C--(0.15,0.99)); draw((1.03,0.34)--T); draw(D--(0.15,0.99)); draw(C--F); draw(A--Q); draw(B--Q); draw(A--P); draw(P--B); draw(C--D); draw(D--G); draw(G--C); draw(A--B); draw(D--A); draw(C--B); draw(G--I); draw(J--A); draw(J--G); draw((0.15,0.99)--J); /* dots and labels */ dot(P,uququq); label("$P$", (0.02,0.03), NE * labelscalefactor,uququq); dot(A,xdxdff); label("$A$", (0.77,0.68), NE * labelscalefactor,xdxdff); dot(B,xdxdff); label("$B$", (0.9,-0.44), NE * labelscalefactor,xdxdff); dot(Q,uququq); label("$Q$", (1.23,0.16), NE * labelscalefactor,uququq); dot(C,xdxdff); label("$C$", (1.28,0.85), NE * labelscalefactor,xdxdff); dot(D,xdxdff); label("$D$", (1.17,-0.53), NE * labelscalefactor,xdxdff); dot((0.15,0.99),uququq); label("$E$", (0.17,1.02), NE * labelscalefactor,uququq); dot(F,uququq); label("$F$", (0.45,-0.87), NE * labelscalefactor,uququq); dot((1.03,0.34),uququq); label("$S$", (1.05,0.37), NE * labelscalefactor,uququq); dot(T,uququq); label("$T$", (0.96,-0.42), NE * labelscalefactor,uququq); dot(G,uququq); label("$G$", (-0.98,0.11), NE * labelscalefactor,uququq); dot(H,uququq); label("$H$", (0.72,-0.13), NE * labelscalefactor,uququq); dot(I,uququq); label("$I$", (1.06,-0.08), NE * labelscalefactor,uququq); dot(J,uququq); label("$J$", (-0.79,2.49), NE * labelscalefactor,uququq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] ez Note that $DA \perp GC$ and $BC \perp GD$ so we get that $I$ is the orthocenter of $\triangle GCD$ so it also lies on the circle $\alpha$. now let the tangent at $I$ to $\alpha$ intersect the line $BQ$ at $T'$. now by applying pascal and duality lemma we get that $B-T'-H-E-J$ are collinear so $T'=T$. doing the same thing for $S$ we get that $S$ also lies on the tangent from $I$ to $\alpha$ so $ST$ is tangent to $\alpha$. now we are done since $GI$ is perpendicular to both $\overline{ST}$ and $\overline{CD}$.$\blacksquare$ Another idea: there could be another way using $\textit{moving points}$ and $\textit{steiner conic}$ theorem but i wasn't able to complete that idea, unfortunately.

here is a proof using mostly pole-polar: lemma 1: let $ABCD$ be a cyclic quadrilateral and let $AD\cap BC=E, AC\cap BD=F, AA\cap BB=X, CC\cap DD=Y$ where by $AA$ we mean the tangent to the circle at point $A$. then $EFXY$ are collinear. proof: using pascal's theorem in $AADBBC$ and $DDACCB$ we get $E, X, F$ and $E, Y, F$ are collinear so $EXYF$ are collinear. back to the problem: define $\rho (X)$ the polar line of $X$ with respect to $\alpha$. we know that $D \in \rho (C)$ because the midpoint of $DC$ is on the radical axis of $C,\alpha$ (because $QB^2=QC^2$). similarly $C\in \rho (D)$. because we know that $CE,DF$ are tangent to $\alpha$, we get that $\rho (C)=DE, \rho (D)=CF$ let $Y=CB\cap AD$. with simple angle chasing we conclude that $Y$ lies on $/\alpha$. on the other hand, $PY$ is perpendicular to $CD$ as a direct consequence of lemma 1 for points $ACDB$. now: $$T=\rho (B)\cap\rho (C) \Rightarrow BC=\rho (T)$$$$S=\rho (A)\cap\rho (D) \Rightarrow AD=\rho (S)$$$$\Rightarrow Y=AD\cap BC=\rho (T)\cap \rho (S) \Rightarrow ST=\rho (Y)$$so because $Y\in\alpha$, $\rho (Y)$ is tangent to $\alpha$ and perpendicular to $PY$ (thus parallel to $CD$) so $ST$ is tangent to $\alpha$ and parallel to $CD$.

Again, this problem is from American Mathematical Monthly, 12245. See 2021 Issue 4 here. worth noting that it's currently ongoing.

Taking the problems from some source without saying them is wrong ?

UKR3IMO wrote: Taking the problems from some source without saying them is wrong ? You’d better change “?” with “.”

Rickyminer wrote: Again, this problem is from American Mathematical Monthly, 12245. See 2021 Issue 4 here. worth noting that it's currently ongoing. The problem is from Iran Tst 2018. and the magazine you are refering is from 2021. So how about just enjoying the problem? All i see is un needed comments about where the problem is from and i dont know how this helps this tread. If you want a source its from Iran Tst 2018 Day 1 .

You are right. I was writing these things just for your information. As you can see, 2 out of 9 problems are spotted to be from AMM. And this one has the exactly same wording with that AMM problem, making it suspicious. Then it turns out that AMM problem is copied from Iranian TST problem, and Iranians used it again. This is stranger. If you do find the discussion here useless, I shall request deleting them and lock the whole thread, since itself is a duplicate one.

Solved with p_square. Let $PC\cap \beta=X, PD\cap \beta=Y, BC\cap AD=Z$. Claim 1: $Z\in \alpha$ Proof: Observe that we want $\angle BCD+\angle ADC=\angle AEB$ but we have $2\angle BCD+2\angle ADC=180-\angle BQA$ and $2\angle AEB=\angle ABP$ but $ABPQ$ is cyclic so we are done. Claim 2: $X,E,D$ collinear. Observe that $\triangle PEC$ is right angled and $PX\cdot PC=PE^2$. Thus, $\angle EXC=90$. Let $EX\cap \beta=D'$ then $\angle CXD'=90$ and thus $D\equiv D'$. Claim 3: $Y,F,C$ collinear. Proof: Same as claim 2. Claim 4: $BTPZX$ cyclic. Proof: $\angle TBP=\angle TXP=90$, thus $TBXP$ cyclic. Also, $PB^2=PZ^2=PX\cdot PC$ and $B,Z,P$ collinear. Thus, $P,B,Z,X$ cyclic. Thus, the claim follows. Now, we get that $\angle TZP=90$ and similarly $\angle SZP=90$. So we get the desired tangency. Now, we just want $TZ\perp CD$ for the parallel condition. Let $AC\cap BD=Z'$. Now, as before $Z'\in \alpha$. Also, $\angle ZBZ'=90$. Thus, $ZZ'$ is diameter of $\alpha$ and thus $P\in ZZ'$. Thus, in cyclic quadrilateral $ABCD$, let $AB\cap CD=R$. Now, we have $QR\perp ZZ'$($Q$ is center of $(ABCD)$) but $QR=CD$ and $ZZ'=PZ$ and we are done. Very nice problem!!