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 $$\vec{u}=\vec{OC}$$, $$\vec{v}=\vec{OB}$$ and $$\vec{w}=\vec{OA}$$.

If $$AC$$ and $$BC$$ are perpendicular, then $$(\vec{u}-\vec{w})\cdot (\vec{u}-\vec{v})=0$$.

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

Our expression then becomes

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

From the distributive property of the dot product,

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

From the commutivity of the dot product, we get

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

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

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

The result follows.