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Given: P is an external point to the circle C(O,r).
PQ and PR are two tangents from P to the circle.
To Prove: PQ = PR
Construction: Join OP
Proof:
{tex}\because{/tex} A tangent to a circle is perpendicular to the radius through the point of contact
{tex}\therefore{/tex} {tex}\angle{/tex}OQP = 90o = {tex}\angle{/tex}ORP
Now in right triangles POQ and POR,
OQ = OR [Each radius r]
Hypotenuse. OP = Hypotenuse. OP [common]
{tex}\therefore{/tex} {tex}\triangle{/tex}POQ {tex}\cong{/tex} {tex}\triangle{/tex}POR [By RHS rule]
{tex}\therefore{/tex} PQ = PR
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