Question

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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

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Answer 1

Correct Question:

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°.

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Given:

• We have been given that 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.

To Prove:

• We need to prove that ∠AOB = 90°.

Construction:

• Join OC

Solution:

We know that the tangents drawn to a circle from an external point are equal.

Therefore, AP = AC_____(1)

Now, in ΔPAO and ΔCAO, We have

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC [From equation 1]

=> ΔPAO ≅ ΔAOC by SSS congruency criteria.

∴ ∠PAO = ∠CAO

=> ∠PAC = 2∠CAO_____(2)

Similarly, we have ∠CBQ= 2∠CBO__(3)

We know that that sum of angles on the same side of transversal is 180°.

∴ ∠PAC + ∠CBQ = 180°

=> 2∠CAO + 2∠CBO = 180° [From 2 & 3]

=> ∠CAO + ∠CBO = 180/2 = 90°____(4)

Now, in ΔAOB, we have

∠BAO + ∠ABO + ∠AOB = 180° [Angle sum property of a triangle]

From equation 4, we have

∠CAO + ∠CBO + ∠AOB = 180°

=> 90° + ∠AOB = 180°

=> ∠AOB = 180° - 90°

=> ∠AOB = 90°

Therefore, ∠AOB = 90°.

Hence proved!!

Answer 2

Answer:

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