Length of Tangents to a Circle: it is the distance of the point P to the point of contact on the circle that is point A and Point B. Or in other words PA and PB are the lengths on tangent from point P to the circle
Theorem: The length of tangents drawn from an external point to a circle are equal. Given :A circle with centre O. PA and PB to tangents from point P. To Prove: PA = PB Proof: Join OA an OB In and OA = OB [radii of the same circle] OP = OP [ common ] [ each ] Thus [ RHS criterion] PA = PB [CPCT] Thus, lengths of tangents from an external point to a circle are equal. Corollary: Theline joining the centre of the circle and the external point i.e. OP is the angle bisector of AOB and APB. As , [By CPCT] is the angle bisectors of Similarly OP is the angle bisector of AOB |
ILLUSTRATION : A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD =AD + BC
Solution A quadrilateral ABCD circumscribes a cricle. Since the lengths of tangents from an external point to a circle are equal, we get AP = AS [ lengths of tangents from point A ] BP = BQ [ lengths of tangents from point B ] CR = CQ [ lengths of tangents from point C ] DR = DS [ lengths of tangents from point D ] Adding the above expressions, we get AP + BP + CR + DR = AS + BQ + CQ + DS ( AP + BP ) + ( CR + DR ) = ( AS + DS ) + ( BQ + CQ ) AB + CD = AD + BC Hence Proved |
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