I don't think you can do that, because the arms of the inner isosceles can be spread anywhere, so the angle at the top will vary. In the extreme case, if the inner triangle's arms are brought together they will form a perpendicular with an angle at the top of effectively zero and the angle at A bisected by the perpendicular. If you draw the picture again so that CAD and EAB are smaller, the angle DAE will be larger and clearly not equal to the other angles. Perhaps there's more info that hasn't been included, like, for example, trisection of CB, CD=DE=EB?