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