Q. 94.5( 484 Votes )
In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.
Let us join point O to C.
In ΔOPA and ΔOCA,
OP = OC (Radius of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
ΔOPA ≅ ΔOCA (SSS congruence criterion)∠POA = ∠COA … (i)
Similarly, ΔOQB ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°
From equations (i) and (ii), it can be observed that 2∠COA + 2 ∠COB = 180°∠ COA + ∠ COB = 180° / 2
∠COA + ∠COB = 90°
∠AOB = 90°Hence Proved.
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