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

Explanation

We know that the tangents drawn from an external point to a circle are equal.
`:. AD=AF`, …….`(i)` [tangents from `A`]
`BD=BE`, ……….`(ii)` [tangents from `B`]
`CE=CF`………`(iii)` [tangents from `C`]
Now, `AB=AC` [given]
`impliesAD+BD=AF+CF`
`impliesBD=CF`
`impliesBE=CE`[ using `(ii)` and `(iii)`]
`impliesE` bisects `BC`.