In my figure $A$ was closer to $X$. Let $O_1,O_2$ be the centers of $\omega_1,\omega_2$ respectively. Let $SO_2$ intersect $AB$ at $P$. Now,$\angle O_1XT=\angle O_1TX=\angle O_2TS=\angle O_2ST\Rightarrow SO_2\parallel O_1X$. But $O_1X\perp AB$. Thus, $SO_2\perp AB\Rightarrow SA=SB$. Now $\angle XTA=180^{\circ}-\angle ATS=\angle SBA=\angle SAB=\angle STB\Rightarrow \angle STA=\angle BTX$. But we also have $\angle XBT=\angle ABT=\angle AST$ so $\triangle XBT, \triangle ATS$ are spirally similar. Thus, $\angle BXT=\angle SAT=\angle SBT\Rightarrow SA,SB$ both are tangents to the circumcircle of $\triangle ATX$. Again, $\angle BXT=\angle AXT=\angle TYX=\angle TYI$ and $\angle SAT=180^{\circ}-\angle SCT=\angle TCY$. Since $\angle BXT=\angle SAT$ we have $\angle TYI=\angle TCI$ and thus $C,T,I,Y$ are con-cyclic. Let the bisector of $\angle BCA$ be $CQ$. Then $\angle SCQ=\angle SCB+\angle BCQ=\angle SAB+\angle ACQ=\angle SBA+\angle ACQ$$=180^{\circ}-\angle SCA+\angle ACQ=\angle ICA+\angle ACQ=\angle ICQ$. But $\angle SCQ+\angle ICQ=180^{\circ}$ so $\angle ICQ=90^{\circ}$. Since $CQ$ was the internal bisector of $\angle BCA$, $CI$ must be the external bisector. Again, $\angle SIT=\angle CIT=\angle CYT=\angle YXT=\angle IXT$ i.e. $SI$ is tangent to the circumcircle of $\triangle TIX$. So, $SI^2=SX\times ST=SA^2=SB^2$ since $SA,SB$ both were tangents to the circumcircle of $\triangle ATX$. Thus $S$ is the circumcenter of $\triangle ABI$. So, $\angle ABI+\angle CBI=\angle ABC=\angle ASC=\angle ASI=2\angle ABI$. Thus we have $\angle ABI=\angle CBI\Rightarrow BI$ is the internal bisector of $\angle CBA$. So, $I$ is the $B$-excenter of $\triangle ABC$.

Why is <ASI = 2<ABI?

Dear Mathlinkers, for (a) we can applied a converse of the Miquel's theorem (pivot). Sincerely Jean-Louis

fmasroor wrote: Why is <ASI = 2<ABI? Because $S$ is the center of the circle $ABI$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3181868&sid=10ecee2054893c454bd1f0269924a031#p3181868

Let $\angle{TSC}=x,\angle{TXY}=y,\angle{TYX}=z,\angle{TCA}=w$. Part I: $\angle{CIY}=\angle{CTY}=x+y$ so $TCYI$ is cyclic. Part II: Thus we have $\angle{ICA}=\angle{ICT}+\angle{TCA}=\angle{IYT}+\angle{TCA}=z+w$.$\angle{ABS}=\angle{ABT}+\angle{TBS}=\angle{ACT}+\angle{TCI}=w+z$.Thus from $\triangle{XSB}$ we get $\angle{TSB}=\angle{XSB}=180-2z-w=\angle{TCB} \implies \angle{ACB}=180-2z-2w$.Thus $I$ lies on the external bisector of $\angle{ACB}$.Now note that $\angle{SXI}=\angle{SIT} \implies \triangle{SXI} \sim \triangle{SIT} \implies ST \cdot SX=SI^2$.Now also note that $\angle{TAS}=\angle{TBS}=z==\angle{SXA} \implies ST \cdot SX=SA^2$ similarly.Thus $SI=SA$ and $\angle{ISA}=x+w$.Thus $\angle{SIA}=90-\frac{x+w}{2}=90-\frac{\angle{ABC}}{2}$ conforming that $I$ is the excenter,as desired.

Claim: $SA=SB$. Proof: Through a homothety at $T$, we have $\overline{XO_2} \parallel \overline{SO_1}$, so both are perpendicular to $\overline{AB}$. The result follows. $\blacksquare$ So it follows that $\overline{CI}$ externally bisects $\angle SCB$. Claim: [Part (a)] $CTYI$ is cyclic. Proof: Angle chase! \[\measuredangle TCI = \measuredangle TCB + \measuredangle BCI = \measuredangle TAB + \measuredangle BAS = \measuredangle TAS = \measuredangle TYX = \measuredangle TYI\]by the same homothety at $T$. $\blacksquare$ Claim: $S$ is the circumcenter of triangle $ABI$. Proof: Our cyclic quadrilateral yields $\measuredangle TIS = \measuredangle TYC = \measuredangle TXY$, so triangles $TSI$ and $ISX$ are similar. Similarly, $\measuredangle BTS = \measuredangle XBS$, so triangles $TSB$ and $BSX$ are similar. Thus $SI^2 = ST \cdot SX = SB^2$, and we have $SA=SB$ from the first claim, as needed. $\blacksquare$ So $\angle IAC = \frac 12 \angle BSC = \frac 12 \angle BAC$, and we're done!