Thales' Theorem

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

Proof


Let $O$ be the centre 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.