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: ABC is an isosceles triangle in which AB = AC; with a circle inscribed
in the triangle.
To prove: BD = DC
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
We also know that Δ ABC is an isosceles triangle in which AB = AC.
a+b = a+c
Thus b = c
Therefore, BD = DC
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