In the diagram A, B, and C are the points on the circle. Use the diagram to prove the theorem which states that: if O is the center of the circle then AÔB = 2Ĉ.
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Explanation:

Here's one way to go at it.

Draw segments AB and CO. Define angles as follows. (The triangles with sides that are radii are all isosceles, so their base angles are congruent.)

  x = angle OAB = angle OBA

  y = angle OAC = angle OCA

  z = angle OBC = angle OCB

Consider the angles at each of the points A, B, C.

At A, we have ...

  angle CAB = x + y

At B, we have ...

  angle CBA = x + z

At C, we have ...

  angle ACB = y + z

The sum of the angles of triangle ABC is 180°, as is the sum of angles in triangle ABO. This gives ...

  x + x + ∠AOB = (x+y) +(x+z) +(y+z)

  ∠AOB = 2(y+z) = 2∠ACB

This shows ∠AOB = 2×∠C, as required.

this doesn't make sense lol