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[parent] proof of Thales' theorem (Proof)

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

\includegraphics{thales.eps}

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

$\displaystyle \angle ACB=\angle MCB+\angle ACM=90^\circ.$
QED.



"proof of Thales' theorem" is owned by mathwizard.
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Cross-references: QED, isosceles, triangles, circle, center
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This is version 3 of proof of Thales' theorem, born on 2002-06-06, modified 2002-06-09.
Object id is 3064, canonical name is ProofOfThalesTheorem.
Accessed 11485 times total.

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AMS MSC: 51-00 (Geometry :: General reference works )
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