Two tangent segments PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP.

Question

Two tangent segments PA and PB are drawn to a circle with centre O such that \(\angle APB=120^\circ\) Prove that OP = 2 AP.

Answer 1

Given: Two tangent segments PA and PB are drawn to a circle with centre O such that \(\angle APB=120^\circ.\)

To prove: OP = 2 AP

Proof: Construct the figure according to the conditions given.

HereIn triangle OAP and OBP,PA 

= PB (Length of Tangents from external point are equal)

OA = OB (Radii of same circle )

OP = OP (common)

Δ OAP ∼ Δ OBP ( By SSS criterion)

∠OPA = ∠OPB = \(60^\circ\)

In Triangle OAP , ∠OAP = \(90^\circ\)  (By theoram which states that tangent to a circle is perpendicular to the radius through the point of contact)We know in a right angle triangle

⇒ sin 60o= AP / OP , i.e 1/2 = AP / OP

So,OP = 2 AP

Hence proved.