Question

The incircle of an isosceles triangle $ABC,BA=BC$, touches the sides $AB,BC$ and $CA$ at $D,E$ and $F$ respectively. Prove that $F$ bisects $AC$.

Answer 1

Given : $BA=BC$ ---(1)

$BD=BE$ (Tangents from point $B$) ---(2)

On subtracting equation (1) and (2), we get

$BA-BD=BC-BE$

$DA=EC$ ---(3)

$EC=CF$

$CF=AD$----(4)

$AD=AF$----(5)

Therefore, from equation (4) and (5),

$CF=AF$

Hence, $F$ bisects $AC$.