Properties of circle................

Join OC. As OB = OC = Radius of circle, so $\angle OBC = \angle OCB = 70 ^ \circ$ . Applying angle sum property in $\triangle$ BOC, we get $\angle BOC = 40 ^ \circ$ .

Now, $\angle BOC = \angle COD = 40 ^ \circ$ {As angle formed by equal chords on the center are equal}.

$\angle BAC = \frac{1}{2}\angle BOC = \frac{1}{2}40 ^ \circ = 20 ^ \circ$ {As angle made by a chord on the circumference is half the angle made by it on the center}.

$\angle BAC = \angle CED = 20 ^ \circ$ {As angle formed by equal chords on the circumference are equal}.

$\angle BFD = \frac{1}{2}\angle BOD = \frac{1}{2}80 ^ \circ = 40 ^ \circ$ . {As angle made by a chord on the circumference is half the angle made by it on the center}

We have to find $\angle BAC + \angle BFD + \angle CED$ which is equal to $20^ \circ + 40^ \circ + 20^ \circ$ = $\boxed{80^\circ}$ and that's our answer.