InterviewSolution
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InterviewSolution
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1 In the Fig. 2, given below, OB is the perpendicular bisector of the line segment DE, FA LOB andFE intersects OB at the point C. Prove thatROA OB OC |
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Answer» OB is perpendicular bisector of line segment DE , FA perpendicular to OB and FE intersects OB at the point C as shown in figure . now, ∆OAF and ∆ODB ∠OAF = ∠OBD = 90° {because OB is perpendicular bisector of DE so, OB ⊥ DE and OB ⊥ AF } ∠FOA = ∠DOB { common angle } from A - A similarity , ∆OAF ~ ∆ODB so, OA/OB = AF/DB = OF/OD -------(1) similarly, ∆AFC and ∆BEC ∠FCA = ∠BCE ∠FAC = ∠CBE = 90° from A - A similarity , ∆AFC ~ ∆BEC so, AF/BE = AC/CB = FC/CE -------(2) we know, DB = BE { perpendicular bisector of DE is OB } put it in equation (2) AF/DB = AC/CB =FC/CE -------(3) now, from equations (1) and (3), OA/OB = AC/CB = ( OC - OA)/(OB - OC) OA/OB = (OC - OA)/(OB - OC) OA(OB - OC) = OB(OC - OA) OA.OB - OA.OC = OB.OC - OB.OA 2OA.OB = OB.OC + OA.OC dividing by OA.OB.OC both sides, 2OA.OB/OA.OB.OC = OB.OC/OA.OB.OC + OA.OC/OA.OB.OC2/OC = 1/OA + 1/OB |
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