What's the measure of the segment $UQ$ in the circumference below?

Question

For reference: On the circumference of diameter $AB$, chord $PQ$ is drawn perpendicular to $AB$ at point $R$ ; on the arc $AQ$ take the point $U$ such that $PU$ intersects $AB$ and $AQ$ at $S$ and $T$ respectively, if $PS.TU= 8TS$, calculate $UQ$.(Answer:8)

My progress;

PAUQ is cyclic $\implies \angle QAP = \angle QUP,\\ \angle AQU = \angle APU\\ \angle UPQ = \angle UAQ$

Intersecting Chords Theorem: $PS.SU = AS.SB\implies\\ PS.(ST+TU)=AS.SB \rightarrow PSST+PSTU = AS.SB \rightarrow PSST+8TS = AS.SB$ $\triangle AQP$ is isosceles $\implies $AQ = AP$

but I can't visualize anything else.

Answer 1

In $\triangle PAT$, $AS$ is angle bisector. So $$\frac{PS}{TS}=\frac{AP}{AT}$$

Also $\triangle TAP \sim \triangle TUQ$ by angle-angle criterion so that $$\frac{AP}{AT}=\frac{UQ}{UT}$$

From above two, $$\frac{PS}{TS} = \frac{UQ}{UT}$$

$$\Rightarrow PS \cdot TU = UQ \cdot TS$$

Conclusion, $UQ=8$.

Answer 2

Here is another way.

As $\angle AQS = \angle APS = \angle AQU$, $QT$ is angle bisector of $\angle SQU$.

So, $ \displaystyle \frac{UQ}{SQ} = \frac{TU}{TS} = \frac{8}{PS}$

But as $SQ = PS, UQ = 8$