In a circle through the middle O of the chord AC, the chord BD is drawn so that the arcs AB and CD

In a circle through the middle O of the chord AC, the chord BD is drawn so that the arcs AB and CD are equal. Prove that O is the midpoint of the chord BD.

Consider triangles AOB and COD.

Segment OA = OC, since point O is the middle of the chord AC, AB = CD by condition.

The inscribed angle BAC rests on the arc BC as does the inscribed angle BDC, then the angle OAB = ODC, and then the triangles AOC and COD are equal on two sides and the angle between them, which means OB = OD, and point O is the middle of the chord BD, which was required prove.