Prove that lengths of tangents drawn from an external point to a circle are equal.

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