Let AngleCBE = AngleDBE = y (BE is angle bisector)
let angleBCE = angleFCE = z (CE is angle bisector)

angleBAC = X = 50

Thus we can form an equation,
angleBAC+angleBCA+angle ABC = 180

angle ABC+y +y = 180

angle ABC = 180-2y
simillary,
angle ACB= 180 - 2z

Thus ,
180-2y+180-2z+50=180
2y+2z=230
y+z=115...(1)

Now triangleBCE
Let angle E be p
so,
angle CBE+angleBCE+CEB = 180
i.e, angle E +y + z = 180

(from 1,y +z = 115)
thus angle E = 180-115
=65
thusD

thanks.