Let us assume AB and AC are tangents to a circle with center O. We need to prove that AO bisects the chord BC at a right angle. Proof: AB and AC are tangents to the circle with center O. Therefore, AB = AC And AO is the bisector of \(\angle\)BAC. \(\therefore \angle\)BAD = \(\angle\)CAD In triangles \(\triangle\)ABD and \(\triangle\)ACD: AB = AC \(\angle\)BAD = \(\angle\)CAD And AD is the common side \(\therefore \triangle\)ABD \(\cong \triangle\)ACD (By SAS congruence rule) Hence, BD = CD [corresponding sides] And \(\angle\)BDA = \(\angle\)CDA [corresponding angles] Again, \(\angle\)BDA + \(\angle\)CDA = 180\(^\circ\) Or, 2\(\angle\)BDA = 180\(^\circ\) [\(\because \angle\)BDA = \(\angle\)CDA] \(\therefore \angle\)BDA = 90\(^\circ\) \(\therefore\) AO bisects the chord at a right angle. (Proved)