Q.

In the below figure, 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°.


Ans.

Join OC. Since, the tangents drawn to a circle from an external point are equal.

∴ AP = AC

In Δ PAO and Δ AOC, we have:

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC

⇒ Δ PAO ≅ Δ AOC [SSS Congruency]

∴ ∠PAO = ∠CAO = ∠1

∠PAC = 2 ∠1                              ...(1)

Similarly ∠CBQ = 2 ∠2               ...(2)

Again, we know that sum of internal angles on the same side of a transversal is 180o.

∴ ∠PAC + ∠CBQ = 180o

⇒2 ∠1 + 2 ∠2 = 180o [From (1) and (2)]

⇒ ∠1 + ∠2 = 180o/2=90o                           ...(3)

Also ∠1 + ∠2 + ∠AOB = 180° [Sum of angles of a triangle]

⇒ 90° + ∠AOB = 180o

⇒ ∠AOB = 180o-90o ⇒ ∠AOB = 90o.