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: ΔABC is an isosceles triangle in which AB=AC, with a circle inscribed in the triangle.
To prove: BD=DC
Proof: AF and AE are tangents drawn to the circle from point A. Since two tangents drawn to a circle from the same exterior point are equal, AF=AE=a
Similarly, BF=BD=b, and CD=CE=c
ΔABC is an isosceles triangle in which AB=AC
So, a+b = a+c
Thus, b=c
Therefore, BD=DC
Hence proved.