Thales' Theorem/Proof 3

Proof
Let $O$ be the center of $ACB$.

We have that $AO$, $BO$ and $CO$ are radii of the same circle.

Thus by definition:
 * $AO = BO = CO$

Thus $\triangle AOC$ and $\triangle BOC$ are isosceles triangles.

From External Angle of Triangle equals Sum of other Internal Angles:
 * $\angle COB = \angle ACO + \angle CAO$

Hence:

Similarly, from External Angle of Triangle equals Sum of other Internal Angles:
 * $\angle COA = \angle BCO + \angle CBO$

Hence:

Therefore:

But $\angle COB + \angle COA$ equals two right angles.

Hence $\angle ACB$ is a right angle.