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Answer 1

Given: XY and X'Y' at are two parallel tangents to the circle wth centre O and AB is the tangent at the point C,which intersects XY at A and X'Y' at B. To prove: ∠AOB = 90º . Construction: Join OC. Proof: In ΔOPA and ΔOCA OP = OC (Radii ) AP = AC (Tangents from point A) AO = AO (Common ) ΔOPA ≅ ΔOCA (By SSS  criterion)Therefore, ∠POA = ∠COA .... (1)   (By C.P.C.T)Similarly , ΔOQB ≅ ΔOCB ∠QOB = ∠COB .......(2) POQ is a diameter of the circle. Hence, it is a straight line. Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180° From equations (1) and (2), it can be observed that 2∠COA + 2∠COB = 180° ∴ ∠COA + ∠COB = 90° ∴ ∠AOB = 90
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