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Question

In the fig. 3.21 seg ABseg AD and seg BC seg DC prove that seg AC is perpendicular bisector of Side BD.

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Solution

Given : seg AB segADOr, AB=ADseg BC seg DCBC=DCIn ABC and ADC,AB =AD GivenBC = DC GivenAC = AC (common)ABC ADC (SSS test)Thus, BAP = DAP (by c.a.c.t) ...(1)In ABP and ADP,AB= AD BAP = DAP from (1)AP = AP (common)ABP ADP (SAS test)Thus, BP = PD (by c.s.c.t) andAPB = APD (by c.a.c.t)Since APB+ APD = 180° linear pair,2 APB = 180° APB = 90°= APD Hence, AC is the perpendicular bisector of BD.

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