# 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 $\angle BMC=:\alpha$ then $\angle MCB=90^{\circ}-\frac{\alpha}{2}$ and $\angle CMA=180^{\circ}-\alpha$. Therefore $\angle ACM=\frac{\alpha}{2}$ and

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

QED.

Title proof of Thales’ theorem ProofOfThalesTheorem 2013-03-22 12:45:27 2013-03-22 12:45:27 mathwizard (128) mathwizard (128) 6 mathwizard (128) Proof msc 51-00