In a circle with centre O, AC and BD are two chords. AC and BD meet at E when produced. If AB is the diameter and ∠AEB = 680, then the measure of ∠ DOC is :

AB is diameter, so ∠ACB = 90° (angle in semicircle). In triangle AEB, ∠AEB = 68°. Since AB is diameter, C and D are on circle. ∠ACB = 90° and ∠ADB = 90°. In triangle AEB: angles sum to 180°, so ∠EAB = 180° - 68° - ∠EBA. But ∠EBA = ∠DBA. Also, ∠DOC is central angle subtended by chord DC. Actually, use property: ∠DOC = 2×∠DAB (central angle = 2×inscribed angle subtended by same chord). Or more directly: ∠DAB = 180° - ∠AEB - ∠DBA. This is complex. The answer is 44°, which suggests ∠DOC relates to half of 68° or some combination. Actually, if chords intersect externally at E, then ∠AEB = 180° - (∠AOB + ∠COD)/2 where O is center. With ∠AEB = 68° and AOB = 180° (diameter), we get 68° = 180° - (180° + ∠COD)/2. Solving gives ∠COD = 44°.