Question

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

Answer 1

Given : AP and AQ are two tangents from a point A to a Circle C(0,r).

To prove : AP=AQ

Construction : Join OP, OQ and OA.

Proof : In olrder to prove that AP= AQ , we shall first prove that$\Delta OPA\cong \Delta OQA$

Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore,$OP\perp AP$ and $OQ\perp AQ$

$\Rightarrow \angle OPA=\angle OQA={90}^{\circ }$ Now, in right triangles OPA and OQA, we have

OP = OQ [radii of circle]

$\angle OPA=\angle OQA$ [Each ${90}^{\circ }$]

OA=OA [Common]

So, by RHS – criterion of congruence, we get

$\Delta OPA\cong \Delta OQA\Rightarrow AP=AQ$ [CPCT]

Hence, lengths of two tangents from an external point are equal.