# Thales' Theorem/Proof 1

## Theorem

Let $A$ and $B$ be two points on opposite ends of the diameter of a circle.

Let $C$ be another point on the circle such that $C \ne A, B$.

Then the lines $AC$ and $BC$ are perpendicular to each other.

## Proof

Let $O$ be the center of the circle, and define the vectors:

- $\mathbf u = \overrightarrow{OC}$
- $\mathbf v = \overrightarrow{OB}$
- $\mathbf w = \overrightarrow{OA}$

Thus:

- $\overrightarrow{AC} = \mathbf u - \mathbf w$
- $\overrightarrow{BC} = \mathbf u - \mathbf v$

Then:

\(\ds \) | \(\) | \(\ds \overrightarrow{AC} \cdot \overrightarrow{BC}\) | where $\cdot$ is the dot product | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\mathbf u - \mathbf w} \cdot \paren {\mathbf u - \mathbf v}\) | from above | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\mathbf u + \mathbf v} \cdot \paren {\mathbf u - \mathbf v}\) | Since $A$ is directly opposite $B$ in the circle, $\mathbf w = - \mathbf v$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf v \cdot \mathbf u - \mathbf v \cdot \mathbf v\) | Dot Product Distributes over Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds \mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf v - \mathbf v \cdot \mathbf v\) | Dot Product Operator is Commutative | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf u}^2 - \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf v - \norm {\mathbf v}^2\) | Dot Product of Vector with Itself | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf u}^2 - \norm {\mathbf v}^2\) | simplification | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf u}^2 - \norm {\mathbf u}^2\) | as $\norm {\mathbf u} = \norm {\mathbf v}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds 0\) |

From Non-Zero Vectors are Orthogonal iff Perpendicular, it follows that $AC$ and $BC$ are perpendicular to each other.

$\blacksquare$

## Source of Name

This entry was named for Thales of Miletus.

## Historical Note

This result was known by the ancient Babylonians as early as $2000$ BCE.

The proof of what is now known as Thales' Theorem that Thales of Miletus actually used is unknown. Whether he was aware of the Interior Angles Theorem and used it in his proof cannot be decided.

The Interior Angles Theorem is traditionally ascribed to Pythagoras, who lived near Thales, and may have met him.

Thus it is possible that Pythagoras learned of the Interior Angles Theorem either directly or indirectly from Thales himself.