I assume that P, Q, R, S, T, U follow on the line in this order. Since QR = ST, the circumcenter O of the triangle $\triangle ABC$ lies on the perpendicular bisector of the segment RS. Let (O') be the circumcircle of the triangle $\triangle ARS,$ the circumcenter O' also lies on the perpendicular bisector of RS. The circumcircles (O), (O') intersect at the common vertex A and at another point A'. $AA' \parallel RS$ (both perpendicular to OO'). Quadrilateral ARSA' is a cyclic trapezoid, isosceles. Hence $\angle BAA' \equiv \angle RAA' = \angle USA'.$ Quadrilateral ABCA' is cyclic, hence $\angle BAA' = \angle UCA'.$ It follows that $\angle USA' = \angle UCA'$ and the quadrilateral UCSA' is cyclic. Let a tangent to the circle (O) at A' meet the line PU at a point U'. Since $AA' \parallel SU',$ $\angle SCA' \equiv \angle ACA' = \angle SU'A'$ and the quadrilateral U'SCA' is also cyclic. But the circumcircle of the triangle $\triangle CSA'$ intersects the line $PU \equiv SU$ at the point S and at one other point, therefore $U' \equiv U$ are identical. The points Q, T and the points A, A' are symmetrical with respect to OO' (the perpendicular bisector of RS). Since AP is a tangent to (O) at A and A'U a tangent to (O) at A', the points P, U are also symmetrical with respect to OO' hence PQ = UT.

use Butterfly theory and Pascal theory to solve this problem, easy