In ΔPQR, the bisector of ∠QPR intersects QR at S and the circumcircle of ΔPQR at T. If PQ : PS = 3 : 5, then PT : PR is ?
Answer & explanation
Correct answer: option 1
Here,
∠PTQ = PRQ, (angle made by same chord PQ)
⇒ ΔPSR ∼ ΔPQT
⇒\(\frac{PS}{PR}\) = \(\frac{PQ}{PT}\)
⇒ \(\frac{5}{PR}\) = \(\frac{3}{PT}\)
= \(\frac{PT}{PR}\) = \(\frac{3}{5}\)