Let AB and CD be two chords of a circle with center O. OM and ON are perpendiculars drawn from the center to the chords such that OM = ON. To prove: AB = CD Construction: Join O to B and O to D. Proof: In triangles \(\triangle OBM\) and \(\triangle ODN\): OM = ON [Given] OB = OD [Both are radii of the circle] \(\angle OMB = \angle OND\) [Both are right angles] \(\therefore \triangle OBM \cong \triangle ODN\) \(\therefore BM = DN\) [Corresponding sides] Also, since a perpendicular from the center of a circle to a chord bisects the chord, AB and CD are both bisected by OM and ON respectively. Hence, AB = CD [Proved]