proof of Thales’ theorem

Let $M$ be the center of the circle through $A$, $B$ and $C$.

Then $AM=BM=CM$ and thus the triangles $AMC$ and $BMC$ are isosceles. If $\mathrm{\angle }BMC=:\alpha$ then $\mathrm{\angle }MCB={90}^{\circ }-\frac{\alpha }{2}$ and $\mathrm{\angle }CMA={180}^{\circ }-\alpha$. Therefore $\mathrm{\angle }ACM=\frac{\alpha }{2}$ and

$$\mathrm{\angle }ACB=\mathrm{\angle }MCB+\mathrm{\angle }ACM={90}^{\circ }.$$

QED.