Thales' Theorem

Theorem
If $A$ and $B$ are two points on opposite ends of the diameter of a circle, and $C$ is another point of the circle such that $C \ne A,B$, then the lines $AC$ and $BC$ are perpendicular to each other.

Proof

 * Thales theorem.jpg

Let $O$ be the center of the circle, and define the vectors $\mathbf u = \overrightarrow{OC}$, $\mathbf v = \overrightarrow{OB}$ and $\mathbf w = \overrightarrow{OA}$.

If $AC$ and $BC$ are perpendicular, then $\left({ \mathbf u - \mathbf w}\right) \cdot \left({\mathbf u - \mathbf v}\right) = 0$ (where $\cdot$ is the dot product).

Notice that since $A$ is directly opposite $B$ in the circle, $\mathbf w = - \mathbf v$.

Our expression then becomes


 * $\left({\mathbf u + \mathbf v}\right) \cdot \left({\mathbf u - \mathbf v}\right)$

From the distributive property of the dot product,


 * $\left({ \mathbf u + \mathbf v}\right) \cdot \left({\mathbf u - \mathbf v}\right) = \mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf v \cdot \mathbf u - \mathbf v \cdot \mathbf v$

From the commutativity of the dot product and Dot Product of a Vector with Itself, we get


 * $\mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf v \cdot \mathbf u - \mathbf v \cdot \mathbf v = \left|{\mathbf u}\right|^2 - \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf v - \left|{\mathbf v}\right|^2 = \left|{\mathbf u}\right|^2 - \left|{\mathbf v}\right|^2$

Since the vectors $\mathbf u$ and $\mathbf v$ have the same length (both go from the centre of the circle to the circumference), we have that $|\mathbf u| = |\mathbf v|$, so our expression simplifies to


 * $\left|{\mathbf u}\right|^2 - \left|{\mathbf v}\right|^2 = \left|{\mathbf u}\right|^2 - \left|{\mathbf u}\right|^2 = 0$

The result follows.

Proof 2
From the Inscribed Angle Theorem, $\angle AOB = 2 \angle ACB$.

Then we have that $\angle AOB$ is a straight angle.

Hence the result.

Legend has it that he sacrificed an ox in honour of the discovery.

On the other hand, some attribute this theorem to Pythagoras.