Figure 2

Since $AS = AT$, we have $\angle AST = \angle ATS = \frac{1}{2}\angle BAC$. Similarly, $\angle ARU = \angle AUR = \frac{1}{2}\angle BAC$, so that $\angle S = \angle R$. This implies that $R, S, T$ and $U$ are concylic.

We now show that the centre $O$ of the circle $\omega$ through $R, S, T$ and $U$, lies on the circumcircle of $ABC$. We know that $O$ lies on the perpendicular bisector of $ST$ (which is also the perpendicular bisector of $RU$, since $ST \parallel RU$). This perpendicular bisector forms a diameter of $\omega$, and contains $A$.

If $O = A$, then we are finished, since $A$ is certainly on the circumcircle of $ABC$. So suppose that the points $O$ and $A$ are different, as shown in Figure 2. Drop perpendiculars from $O$ to $BR$ (which is $BA$ extended), to $AC$, and to $CB$. Let the feet of these perpendiculars be $I, J$ and $K$, respectively. It is known that $O$ is on the circumcircle of $ABC$ if and only if the points $I, J$ and $K$ are collinear. (In case this happens, the line through $I, J, K$ is called the Simson line of $ABC$ determined by $O$.)

In order to show that $I, J, K$ are collinear, it is sufficient to show that $\angle JKO = \angle IKO$. Since $\angle OJC = \angle OKC = 90^\circ$, $OJKC$ is a cyclic quadrilateral, so that $\angle JKO = \angle JCO$. Our next observation is that $IT = JU$. This follows from the fact that $OI = OJ$ (from symmetry – recall that triangle $RAU$ is isosceles, and $AO$ is on the perpendicular bisector of $RU$), and $OT = OU$, giving $IT^2 = OT^2 - OI^2 = OU^2 - OJ^2 = JU^2$. Hence, $IB = IT - BT = JU - CU = JC$, from which we get that triangles $OIB$ and $OJC$ are congruent. We therefore have $\angle IBO = \angle JCO$. Finally, since $IOKB$ is a cyclic quadrilateral ($\angle BIO = \angle BKO = 90^\circ$), we also have $\angle IBO = \angle IKO$.

Putting everything together, we conclude that $\angle JKO = \angle JCO = \angle IBO = \angle IKO$, and we are done.

The same strategy as in Solution 1 shows that $RSTU$ is cyclic.

The centre $O$ must lie on the four perpendicular bisectors of $RS, ST, SU$ and $RU$. But since triangles $RAU$ and $SAT$ are isosceles, the perpendicular bisectors of $ST$ and $RU$ pass through $A$ and bisect the angle $RAU$.

Now $\angle TOU = 2\angle TSU$, since $O$ is the centre of circle $RSTU$, which equals $\angle TSU + \angle ATS$, since $AS=AT$, which in turn equals $\angle TAU$, the exterior angle in triangle $AST$. This proves that $AOUT$ is cyclic and hence $\angle BTO = \angle ATO = \angle AUO = \angle CUO$.

Along with $OT=OU$ (radii) and the given $BT=CU$, we conclude that triangles $BTO$ and $CUO$ are congruent, so $\angle BOT = \angle COU$. Finally $\angle BAC = \angle TAU = \angle TOU = \angle BOC - \angle BOT + \angle COU = \angle BOC$ and thus $ABCO$ is cyclic.

(The problem could also be finished by noting that $O$ lies on the perpendicular bisector of $BC$ and the exterior angle bisector of $A$ and thus lies on the circumcircle, but this assumes some extra knowledge that the above presentation doesn't.)