How to use the angle bisector properties to solve this question?

Question

Given that $TQ$ is an angle bisector, the question asks whether the length of $ST$ is equal to $SQ$ or $(SR+SQ)/2$?

I was able to use the tangent secant theorem ($ST*ST=SR*SP$). However, I couldn't make much use of the fact that $TQ$ is an angle bisector. Please guide if/when free.

Try 1: I assumed $ST$ equal to $SQ$ and found that I need to show that $RQ*SQ=SR*QP$, or that $RQ/QP=SR/SQ=TR/TP=SR/TS$, but I couldn't prove any ratio equal to any other.

Try 2: I assumed $ST$ equal to $(SR+SQ)/2$ and found that I need to show that $RQ*RQ=4*SR*QP$, which seems incorrect to me, so I concluded that $ST$ is probably equal to $SQ$ instead of the other option.

Please help if/when free.

Answer 1

The point to note is that $\angle STR = \angle P$. This is in fact one of the approaches to prove tangent-secant theorem (by showing $\triangle SPT \sim \triangle STR$).

Now to show that $\angle STR = \angle P$,

$\angle TOR = 2 \angle P \ $ ($O$ is the circumcenter)

$ \implies \angle OTR = \cfrac{180^0 - 2 \angle P}{2} = 90^0 - \angle P \ $

As $\angle OTS = 90^0, \angle STR = \angle P$

That leads to $\angle STQ = \angle P + \cfrac{\angle T }{ 2}$

You also know, $\angle SQT = \angle P + \cfrac {\angle T }{ 2} $

That should lead you to the right answer.

Answer 2

Hints:

1. $\angle STP = \angle SRT$
2. $\angle STR + \angle SRT = 180^\circ - \angle S$
3. $\angle STQ = \dfrac{\angle STP + \angle STR}{2}$ by the definition of angle bisector of $\angle PTR$
4. $\angle STQ + \angle SQT = 180^\circ - \angle S$

Answer 3

Given $ST$ is tangent to the circle at $T$, and $TQ$ bisects $\angle PTR$.

$\angle PTQ = \angle RTQ$ ... (given angle bisector)

$\angle TPS = \angle RTS$ ... (angles in the same segment)

$\angle PTQ + \angle TPS = \angle RTQ + \angle RTS$

$\angle TQS = \angle QTS$

$TS = QS$ ... (sides of $∆STQ$ opposite equal angles)