Question

9. In Fig. 4.13, 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 1

Step-by-step explanation:

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Answer 2

Let us join point O to C.

In ∆OPA and ∆OCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

∆OPA = ∆OCA (SSS congruence criterion)

Therefore, P C, A - A, O - O

∠POA = ∠COA................(i)

Similarly, ∆OQB=∆OCB

∠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°