In the figure, segment PQ is the diameter of the circle with center O. The tangent to the tangent circle drawn from point C on it, intersects the tangents drawn from points P and Q at points A and B - Geometry Mathematics 2

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Sum

In the figure, segment PQ is the diameter of the circle with center O. The tangent to the tangent circle drawn from point C on it, intersects the tangents drawn from points P and Q at points A and B respectively, prove that ∠AOB = 90°

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Solution

Given: PQ is the diameter of the circle.

Point P, Q, C are points of contact of the respective tangents.

To prove: ∠AOB = 90°

Construction: Draw seg OC

Proof:
In ∆OPA and ∆OCA,

side OP ≅ side OC    ......[Radii of the same circle]

side OA ≅ side OA   ......[Common side]

side PA ≅ side CA   ......[Tangent segment theorem]

∴ ∆OPA ≅ ∠OCA   .....[[SSS test of congruency]

∴ ∠AOP ≅ ∠AOC   ......[C.A.C.T.]

Let m∠AOP = m∠AOC = x   ......(i)

Similarly, we can prove that ∠BOC ≅ ∠BOQ.

Let m∠BOC = m∠BOQ = y   ......(ii)

m∠AOP + m∠AOC + m∠BOC + m∠BOQ = 180°  .....[Linear angles]

∴ x + x + y + y = 180°    ......[From (i) and (ii)]

∴ 2x + 2y = 180°

∴ 2(x + y) = 180°

∴ x + y = 90°  ......(iii)

Now ∠AOB = ∠AOC + ∠BOC

= x + y   ......[From (i) and (ii)]

∴ ∠AOB = ∠AOC + ∠BOC

= x + y    

∴ ∠AOB = 90°   .....[From (iii)] 

  Is there an error in this question or solution?
Chapter 3: Circle - Q.7

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