Question

2. In Fig., XY and XV 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 Z AOB = 90°.​

Answer 1

${\huge{\frak{AnswEr :}}}$

GiveN :

• XY & X'Y' are 2 parallel tangents to a circle with centre O.

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• XY || X'Y'

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• However, if you clearly observe the diagram you can get that AB is tangent at point C.

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For Proving ∠AOB = 90° you'll have to join the points as OC.

From the figure,

As we know that, Tangent at any point of a circle is ⟂ to the radius through point of contact.

That is, OC ⟂ AB

Hence, ∠ACO = 90 = ∠BCO

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Consider the ΔAOP & ΔAOC

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As length of a tangent drawn from external point to a circle are equal.

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That is, AP = AC

OP = CO [Radius]

OA = OA [Common]

∴ ΔAOP ≅ ΔAOC [SSS rule]

So by CPCT,

∠AOP = ∠AOC

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If you consider the line PQ now,

By Linear Pair of Angles :

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↬ ∠AOP + ∠AOC + ∠BOC + ∠BOQ = 180°

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↬ ∠AOC + ∠AOC + ∠BOC + ∠BOC = 180

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↬ 2∠AOC + 2∠BOC = 180

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↬ 2(∠AOC + ∠BOC) = 180

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↬ ∠AOC + ∠BOC = $\dfrac{180}{2}$

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↬∠AOC + ∠BOC = 90

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That is,

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∴ ∠AOB = 90°

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Hence Proved!

Answer 2

$\boxed{\rm{\orange{Given \longrightarrow }}}$

• A circle with the centre "O".
• XY & X'Y' are 2 parallel tangents to the given circle.
• XY || X'Y'

$\boxed{\rm{\red{To\:Prove\longrightarrow }}}$

• ∠AOB = 90° 

$\boxed{\rm{\pink{Proof \longrightarrow }}}$

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

$\sf\purple{→}$2∠COA + 2∠COB = 180°

$\sf\orange{→}$ ∠COA + ∠COB = 90°

$\sf\pink{→}$ ∠AOB = 90°

$\sf\green{Hence\:Proved!♡}$