Question

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°

Answer 1

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)]