Question

Two tangents PA and PB are drawn to a circle with centre O, such that $\angle APB={120}^{\circ }$ Prove that $OP=AP+BP=2AP$

Answer 1

Given: O is the centre of the circle. PA and PB are tangents drawn to a circle and ∠APB = 120°.

To prove: OP = 2AP

Proof:

In ΔOAP and ΔOBP,

OP = OP (Common)

∠OAP = ∠OBP (90°) (Radius is perpendicular to the tangent at the point of contact)

OA = OB (Radius of the circle)

∴ ΔOAP is congruent to ΔOBP (RHS criterion)

∠OPA = ∠OPB = 120°/2 = 60° (CPCT)

In ΔOAP,

cos∠OPA = cos 60° = AP/OP

Therefore, 1/2 =AP/OP

Thus, OP = 2AP

Hence, proved.