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In figure, seg PQ is a diameter of a circle with center O. R is any point on the circle, seg RS ⊥ seg PQ. Prove that, SR is the geometric mean of PS and SQ. [That is, SR2 = PS × SQ]

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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

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