Answer :

Let the external point be A.

AB and AC are the two tangents from A to the circle with radius O.

__Given:__ AP and AQ are 2 tangents on the circle with radius O.

__To Prove:__ AB = AC

Join, BO, AO and CO.

__Proof:__

__Property:__*The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.*

By above property,

OB⊥AB and OC⊥AC

⇒ ∠OBA = ∠OCA = 90°

Now,

In ∆OAB and ∆OAC,

OB = OC [radius]

∠OBA = ∠OCA = 90°

OA = OA [common]

∴By SAS, ∆OBA ≅ ∆OCA

⇒ AB = AC [By CPCTC]

Hence, Proved

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