Question

Two tangent segments PA and PB are drawn to a circle with centre O such that $\angle APB={120}^{\circ }$. Prove that OP = 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 = $\frac{120°}{2}$ = 60° (CPCT)

In ΔOAP,
cos∠OPA = cos 60° = $\frac{AP}{OP}$
$\therefore \frac{1}{2}=\frac{AP}{OP}$
Thus, OP = 2AP