Let AB and CD be two equal chords on the circumference. If you consider the triangles AOB and COD, where O is the center of the circumference, you have:
- AO = CO, because they're both radii
- BO = DO, for the same reason
- AB=CD, given
So, the two triangles have the same side, and are congruent by the SSS theorem.
This implies that they also have the same angles, and in particular, the angles AOB and COD are the same.
