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

Asked by Topperlearning User 18th February 2014, 3:07 PM

Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.

To prove: PA = PB

Construction: Join OA, OB, and OP.

We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

OA PA and OB PB... (1)

In OPA and OPB:

OAP = OBP  = 90o(Using (1))

OP = PO (Common side)

OA = OB (Radii of the same circle)

Therefore, OPA OPB(RHS congruency criterion)

Therefore, PA = PB(Corresponding parts of congruent triangles are equal)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.

Answered by Expert 18th February 2014, 5:07 PM
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