In the given figure O is the centre of the circle, BC and CD are two equal chords and angle OBC = 70 degrees. You have to find sum of angle BAC + angle BFD + angle CED (in degrees).
Then triangle BOC is isoceles with OC = OB,, so angles OBC and OCB are equal
And central angle COB =180 -2(70) = 40°
So angle BAC is an inscribed angle intersecting the same arc as angle COB...so it measures (1/2) 40 = 20°
Note that arc BCD = 2 (40) = 80°
And angle BFD = (1/2)arc BCD = 40°
And angle CED = angle BAC = 20°
So.....BAC + BCD + CED = 20 + 40 + 20 = 80°