Angles in the Same Segment

Pick a chord AB across a circle. Now pick any point on the arc to one side of it and join it to both ends of the chord. The angle you make there is always the same size, no matter which point on that arc you chose. These are angles in the same segment, and they are always equal.

\angle APB = \angle AQB \quad\text{for any } P,\,Q \text{ on the same arc}

It is really just the angle at the centre theorem in disguise: each of these angles is half of the same central angle \angle AOB, so they must all match.

For a chord AB in a circle with centre O:

any two points P and Q on the same arc give equal angles — \angle APB = \angle AQB;
each such angle is half the central angle on the same arc, so \angle APB = \tfrac{1}{2}\,\angle AOB;
so every angle standing on the same arc, on the same side of the chord, has the same size.

Why it works

Two points on the same arc, two angles on the same chord — and they come out equal. Step through the reason.

Both \angle APB and \angle AQB are half of the single central angle \angle AOB, so \angle APB = \angle AQB. Slide P anywhere along that arc and the angle never changes.