In a right ∆ABC, B=90° and D is the mid-point of AC. Prove that BD=AC.

It is given in the question that,

In right triangle ABC, B = 90o


Also D is the mid-point of AC


AD = DC


ADB = BDC (BD is the altitude)


BD = BD (Common)


So, by SAS congruence criterion


ADB CDB


A = C (CPCT)


As, B = 90o


So, by using angle sum property


A = ABD = 45o


Similarly, BDC = 90o (BD is the altitude)


C = 45o


DBC = 45o


ABD = 45o


Now, by isosceles triangle property we have:


BD = CD and


BD = AD


AS, AD + DC = AC


BD + BD = AC


2BD = AC


BD =


Hence, proved


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