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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