Answer:
Yes
Step-by-step explanation:
You can get there a couple of ways. One makes use of the secant rules that tell you ...
PQ × PR = PS × PT
Substituting for PR and PT, you have ...
PQ × (PQ + QR) = PS × (PS + ST)
PQ² + PQ×QR = PS² + PS×ST
Substituting PQ for PS everywhere, we have ...
PQ² + PQ×QR = PQ² + PQ×ST
Dividing by PQ gives ...
PQ + QR = PQ + ST
and subtracting PQ leads us to the conclusion ...
QR = ST
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Another way to look at it is to draw the chord QS. Then ΔQPS is an isosceles triangle, and the perpendicular bisector of QS bisects ∠P and also goes through the circle center. Then the figure is symmetrical about that diameter secant, making QR ≅ ST.
