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Subsection B.1.1 Thales' Theorem

We want to get at right-angled triangles. A classic construction for this is to draw a triangle inside a circle, so that all three corners lie on the circle and the longest side forms the diameter of the circle. See the figure below in which we have scaled the circle to have radius 1 and the triangle has longest side 2.

Thales theorem states that the angle at \(C\) is always a right-angle. The proof is quite straight-forward and relies on two facts:

  • the angles of a triangle add to \(\pi\text{,}\) and
  • the angles at the base of an isosceles triangle are equal.

So we split the triangle \(ABC\) by drawing a line from the centre of the circle to \(C\text{.}\) This creates two isosceles triangles \(OAC\) and \(OBC\text{.}\) Since they are isosceles, the angles at their bases \(\alpha\) and \(\beta\) must be equal (as shown). Adding the angles of the original triangle now gives

\begin{align*} \pi &= \alpha + (\alpha+\beta) + \beta = 2(\alpha+\beta) \end{align*}

So the angle at \(C = \pi - (\alpha+\beta) = \pi/2\text{.}\)