Angle congruence in a basic two circle Geometry problem

Question

I ran across this proof: A basic geometry problem involving circles and I understand why the two sides of the corresponding triangles are congruent, and I understand the accepted answer on this post on why the opposite angles of inscribed quadrilaterals are congruent. However, I can't wrap my head around how that gives the final angle needed for the SAS congruency in this proof. If someone could clarify this for me I'd greatly appreciate it! I feel like I'm just missing something little.

Answer 1

We are at the point where the accepted answer over there writes $\angle RQU=\angle SQT$.

Let $\angle RQU=x$ and $\angle RQS=y$. Then just by adding and subtracting angles we see that $\angle SQU=\angle RQU-\angle RQS=x-y$ and $\angle UQT=\angle SQT-\angle SQU=x-(x-y)=y$. Thus $\angle RQS=\angle UQT=y$.