Question

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

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.

Answer 2

In the given figure,
OA = OB = radius
OP = OP = common
AP = BP = tangents drawn from an external point
Therefore,
Triangle APO similiar to Traingle BPO
$< apo = < bpo \: ( cpct ) = \frac{120}{2} = 60$
$cos \: 60 = \frac{ap}{op} \\ \frac{1}{2} = \frac{ap}{op} \\ so \\ op = 2ap$
Hence proved.