In the given figure, PQ and RS are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects PQ at A and RS at B. Prove that ∠AOB = 90º
Given: PQ and RS are two parallel tangents to a circle with centre O and AB is a tangent to the circle at a point C, intersecting PQ and RS at A and B respectively.
To prove: ∠AOB = 90º
Proof: Since PA and RB are tangents to the circle at P and R respectively and POR is a diameter of the circle, we have
∠OPA = 90º and ∠ORB = 90º
⇒ ∠OPA + ∠ORB = 180º
⇒ PA || RB
We know that the tangents to a circle from an external point are equally inclined to the line segment joining this point to the centre.
∴ ∠2 = ∠1 and ∠4 = ∠3
Now, PA|| RB and AB is a transversal.
∴ ∠PAB + ∠RAB = 180°
⇒ (∠1 + ∠2) + (∠3 + ∠4) = 180°
⇒ 2∠1 + 2∠3 = 180° [ ∵ ∠2 = ∠1 and ∠4 and ∠3 ]
⇒ 2(∠1 + ∠3)= 180°
⇒ ∠1 + ∠3= 90°
From ∆AOB, we have
∠AOB + ∠1 + ∠3 = 180°
[∵ sum of the ∠s of a triangle is 180°]
⇒ ∠AOB + 90° = 180°
⇒ ∠AOB = 90°
Hence, ∠AOB = 90°
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