MHB How show that the points S, U and A are collinear?

[maxkor]

Circle $\omega$ is described on $ABC$. The tangents to the $\omega$ at points $B$ and $C$ intersect at $T$.
Point $S$ lies on the line $BC$ and $AS \perp AT$. Points $B_1$ and $C_1$ are points of intersection of the circle with a radius of $TB$ and center at the $T$ with a line $ST$. Let $BC_{1}$ and $B_{1} C$ cut at the point $G$, and $BB_{1}$ and $CC_{1}$ at the point $U$.
How show that the points S, U and A are collinear?

My try:
Let O the center of circle $\omega$. $SU \perp TG$. So $2\angle BAC=\angle BOC = 180^{\circ}-\angle BTC=\angle BTB_{1}+\angle CTC_{1}=180^{\circ}-\angle BB_{1}T - \angle TBB_{1}+180^{\circ}-\angle CC_{1}T- \angle TCC_{1}=360^{\circ}-2\angle BB_{1}T-2\angle CC_{1}T=2\left(180^{\circ}-\angle BB_{1}T-\angle CC_{1}T\right)=2\left(180^{\circ}-\angle UB_{1}C_{1}-\angle UC_{1}B_{1}\right)=2\angle B_{1}UC_{1}=2\angle BUC$
thus U lies on the circle ABC. And what next?

 

[Mark44]

Circle $\omega$ is described on $ABC$.

@maxkor, what does "described on ABC mean?"

A drawing of the problem would be very helpful.

The tangents to the $\omega$ at points $B$ and $C$ intersect at $T$.
Point $S$ lies on the line $BC$ and $AS \perp AT$. Points $B_1$ and $C_1$ are points of intersection of the circle with a radius of $TB$ and center at the $T$ with a line $ST$. Let $BC_{1}$ and $B_{1} C$ cut at the point $G$, and $BB_{1}$ and $CC_{1}$ at the point $U$.
How show that the points S, U and A are collinear?

My try:
Let O the center of circle $\omega$. $SU \perp TG$. So $2\angle BAC=\angle BOC = 180^{\circ}-\angle BTC=\angle BTB_{1}+\angle CTC_{1}=180^{\circ}-\angle BB_{1}T - \angle TBB_{1}+180^{\circ}-\angle CC_{1}T- \angle TCC_{1}=360^{\circ}-2\angle BB_{1}T-2\angle CC_{1}T=2\left(180^{\circ}-\angle BB_{1}T-\angle CC_{1}T\right)=2\left(180^{\circ}-\angle UB_{1}C_{1}-\angle UC_{1}B_{1}\right)=2\angle B_{1}UC_{1}=2\angle BUC$
thus U lies on the circle ABC. And what next?