Q12 of 50 Page 1

The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.

Given an isosceles triangle ABC.



Also given AB = AC


So, BF + AF = AE + EC … (1)


From the diagram, AB, BC and CA are tangents to the circle at F, D and E respectively.


BF = BD, AE = AF and CE = CD … (2)


From (1) and (2),


BD + AE = AE + CD


BD = CD


Hence proved.


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