Let $X'$ be the reflection of $X$ in $BC$, define $K,L$ as the intersection of perpendicular from $O$ to $BC$ with $(OBC)$ and $(OAD)$. Claim 1:(Key Claim) $\triangle XLAD \sim \triangle X'KBC$

Now $T : XLAD \rightarrow X'KBC$ so $T : KL \rightarrow X'X$ however these two are parallel lines and therefore $T = X'K \cap XL$ so we're done.

haha Let $P$ be $BC\cap AD$. Let the foot from $O$ to $BC$ be $O'$. Let $OO'\cap (\triangle AOD)=O^*$. We claim that $PTO'O^*$ is cyclic. Indeed, \[\measuredangle O'O^*T=\measuredangle ODT=\measuredangle CDT=\measuredangle CPT\]Since $B,C,P$ collinear, $O,O',O^*$ collinear, and since $T$ is the Miquel point clearly of $ABCD$. Hence, $\angle O^*TP=90^\circ$ by Bowtie. It suffices to prove that $X$ lies on $TO^*$. This is equivalent to showing $\angle XTP=90^\circ$. It is well-known, say from many proofs of ISL 2000 G6, that the Clawson-Schmidt Conjugate of $X$ is $Y$, the intersection of the perpendicular bisectors of $CD$ and $AB$. By http://eckartschmidt.de/STEIN.pdf and https://chrisvantienhoven.nl/ql-items/ql-transformations/ql-tf1, the inversion that swaps the reflection of $A$ across the 1st Steiner Axis with $C$ swaps $X$ and $Y$. Hence, it suffices to prove that $\angle YTO=90^\circ$. Let $M,N$ be the midpoints of $AB$ and $CD$ respectively. Clearly, by perpendicular bisector, we only need to show that $OTNYM$ is cyclic. $OMYN$ is cyclic by opposite right angles. Now, as $T$ is the spiral center, being the miquel point, sending $CD$ to $BA$, as $N,M$ are the respective midpoints we have $\measuredangle TNC=\measuredangle TMB$ hence $TOMN$ is cyclic by Bowtie. Thus we are done.

[asy][asy] import olympiad; import cse5; defaultpen(fontsize(10pt)); size(12cm); pointpen=black+linewidth(2); pair X=(0,0),B=(-0.72,0.85),C=(-1.434,-2.5); pair A=(0.72*1.76,0.85*1.76),D=(1.434*1.76,-2.5*1.76); pair O=extension(A,B,C,D),T=(B*D-C*A)/(B+D-C-A); //filldraw(X--B--C--cycle,purple+opacity(0.1),purple); //filldraw(X--A--D--cycle,purple+opacity(0.1),purple); D(A--O--D,purple); pair P=2*foot(circumcenter(O,A,D),T,X)-T; pair Q=2*foot(circumcenter(O,B,C),O,P)-O; D(P--O,purple+dashed); D(circumcircle(O,B,C),magenta); D(circumcircle(O,A,D),magenta); pair OO=circumcenter(O,A,D); pair Xp=reflect(B,C)*X; filldraw(T--Xp--C--Q--B--cycle,invisible,heavygreen); filldraw(T--X--D--P--A--cycle,invisible,blue); D(X--P,blue+dashed); D(Xp--Q,heavygreen+dashed); D(B--C,purple); D(A--D,purple); D(X--Xp,purple+dashed); D("X",D(X),dir(58)); D("A",D(A),unit(A-OO)); D("B",D(B),1.2*dir(67.5)); D("C",D(C),1.1*dir(-90)); D("D",D(D),unit(D-OO)); D("O",D(O),dir(180)); D("T",D(T),dir(90)); D("P",D(P),unit(P-OO)); D("X'",D(Xp),dir(220)); D("Q",D(Q),1.1*dir(289)); [/asy][/asy] Let $X'$ be the reflection of $X$ over $BC$. As $T$ is the Miquel point of quadrilateral $ABCD$, we have $\triangle TCB \stackrel{+}{\sim} \triangle TDA$. Additionally, $\triangle XCB \stackrel{-}{\sim} \triangle XDA$, so $\triangle X'CB \stackrel{+}{\sim} \triangle XDA$, which implies $TX'CB \stackrel{+}{\sim} TXDA$; thus let $\tau$ be the spiral symmetry bringing $TX'CB$ to $TXDA$. Let $P = TX \cap (OAD)$ and $Q = TX' \cap (OBC)$. As $\tau(T) = T$, $\tau(X') = X$, $\tau((OAD)) = \tau((OBC))$, we have $\tau(Q) = P$. Now \begin{align*} \tau(\triangle TX'Q) = \triangle TXP &\implies \triangle TX'Q \stackrel{+}{\sim} \triangle TXP \\ &\implies \triangle TX'X \stackrel{+}{\sim} \triangle TQP \end{align*}Coupled with the fact that $T$, $X'$, $Q$ and $T$, $X$, $P$ are collinear, this implies $PQ \parallel X'X \perp BC$. Thus it suffices to prove that $O$, $P$, $Q$ are collinear. Claim: $O$, $L$, $\tau(L)$ are collinear for any point $L \in (OBC)$. Proof. Let $L' = OL \cap (OAD) \neq O$. Note that \[\measuredangle TLL' = \measuredangle TLO = \measuredangle TBO = \measuredangle TBA\]and \[\measuredangle TL'L = \measuredangle TL'O = \measuredangle TAO = \measuredangle TAB\]which implies $\triangle TLL' \stackrel{+}{\sim} \triangle TBA$; therefore, $\tau(L) = L'$, as desired. $\blacksquare$ Thus we're done.