There are many ways to do this Q. Drongoski pointed out one way but I have a different way
![](https://latex.codecogs.com/png.latex?\bg_white \angle{BCA}=\angle{EBC})
alternate angles on
![](https://latex.codecogs.com/png.latex?\bg_white \parallel)
lines 1
![](https://latex.codecogs.com/png.latex?\bg_white \angle{BCA}=\angle{BAC})
angles in an isosceles triangle. 2
![](https://latex.codecogs.com/png.latex?\bg_white \angle{BCA}+\angle{BAC}=\angle{CBD})
external angle is the sum of two internal angles. 3
Using 1 and sub it into 3 we will have
![](https://latex.codecogs.com/png.latex?\bg_white \angle{BAC}=\angle{EBD})
As such we will see that BE bisects
![](https://latex.codecogs.com/png.latex?\bg_white \angle{CBD})