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1. ABC is an isosceles triangle, in which AB -AC, circumscribed about a crcle Showthat BC is bisected at the point of contact |
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Answer» As tangents drawn from an external point to a circleare equal in length So, therefore, we get, AP=AQ (tangents from A) 1) BP=BR (tangents from B) 2) CQ=CR(tangents from C) 3) As it is given that ABC is an isosceles triangle with sides AB=AC Subtracting AP from both sides, we have, AB-AP=AC-AP implies AB-AP=AC-AQ (from 1) BP=BQ implies BR=CQ (from 2) implies BR=CR(from 3) So therefore BR=CR that shows that BC is bisected at the point of contact. Let the circle touches the side AB at P and side AC at Q and side BC at RWe know that Tangents drawn from external points are equal.Then we have Tangents from point A i.e AP = AQ , Tangents from point B gives BP = BR , Tangents from point C gives RC = CQ.We have AB=AC ⇒ AP+PB=AQ +QC as AP= AQ ⇒ PB = QC ⇒ BR = RCThis gives that BC is bisected at point of contact. |
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