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.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA PA and OB PB
OAP = OBP = 90o ... (1)
In OPA and OPB,
OAP = OBP (Using (1))
OA = OB (Radii of the same circle)
OP = PO (Common side)
Therefore, OPA OPB (RHS congruency criterion)
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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