Question

In Δ ABC,AB = AC If the interior circle of ΔABC touches the sides AB, BC and CA at D, E and F respectively. Prove that E bisects BC.

Answer 1

\(\therefore\) AF = AD as length of tangents from external points are equal.

So,

[AF + FC = AC] = [AD + DB = AB]

Now, 

AF = AD

FC = DB

So, BE = EC

So, (E bisect BC).

Answer 2

AF = AD

BE = BD,

(Tangents from external points)