ABC is an acute angled triangle. $\displaystyle \angle BCA=80 $^{\circ}$$. The angle bisector of $\displaystyle \angle BCA$ meets AB in D. Let the perpendicular bisector of side AB meet the angle bisector of $\displaystyle \angle BDC$ in E. Join E and C. Its given that CE bisects exterior $\displaystyle \angle DCA$ (i.e. $\displaystyle \angle DCE=70$^{\circ}$$).

Prove that $\displaystyle \angle CBA=20$^{\circ}$$

The figure i have attached is not very clear but you do not need the figure. the text i have written is sufficient to understand the question.