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B.1 Theorems about Triangles

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


    Since trigonometry, at its core, is the study of lengths and angles in right-angled triangles, we must include a result you all know well, but likely do not know how to prove.


    The lengths of the sides of any right-angled triangle are related by the famous result due to Pythagoras

    \begin{align*} c^2 &= a^2+b^2. \end{align*}

    There are many ways to prove this, but we can do so quite simply by studying the following diagram:


    We start with a right-angled triangle with sides labeled \(a,b\) and \(c\text{.}\) Then we construct a square of side-length \(a+b\) and draw inside it 4 copies of the triangle arranged as shown in the centre of the above figure. The area in white is then \(a^2+b^2\text{.}\) Now move the triangles around to create the arrangement shown on the right of the above figure. The area in white is bounded by a square of side-length \(c\) and so its area is \(c^2\text{.}\) The area of the outer square didn't change when the triangles were moved, nor did the area of the triangles, so the white area cannot have changed either. This proves \(a^2+b^2=c^2\text{.}\)

    This page titled B.1 Theorems about Triangles is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.