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

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

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

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

Our expression then becomes

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

From the distributive property of the dot product,

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

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

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

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

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

The result follows.

$\blacksquare$

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.

$\blacksquare$

Source of Name

This entry was named for Thales. Legend has it that he sacrificed an ox in honour of the discovery.

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