Question

Solve it fast.......................................

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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