Given: seg PQ is the diameter. seg RS ⊥ seg PQ
To prove: SR2 = PS × SQ
Construction: Extend ray RS, let it intersect the circle at point T.
Proof: seg PQ ⊥ seg RS [Given]
∴ seg OS ⊥ chord RT [R – S – T, P – S – O]
∴ segSR = segTS (i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
Chords PQ and RT intersect internally at point S.
∴ SR × TS = PS × SQ [Theorem of internal division of chords]
∴ SR × SR = PS × SQ [From (i)]
∴ SR2 = PS × SQ