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°.

Answer :

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 = 90°

∠AOB = 90°

Hence Proved.Rate this question :

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PREVIOUSA quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that
AB + CD = AD + BC
NEXTProve that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

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