Thales' Theorem

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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.

Thales-Theorem.png


Proof 1

Thales theorem.jpg

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.

$\blacksquare$


Proof 2

Let $O$ be the center of $ACB$.

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

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

Hence the result.

$\blacksquare$


Proof 3

Let $O$ be the center of $ACB$.

We have that $AO$, $BO$ and $CO$ are radii of the same circle.

Thus by definition:

$AO = BO = CO$

Thus $\triangle AOC$ and $\triangle BOC$ are isosceles triangles.


From External Angle of Triangle equals Sum of other Internal Angles:

$\angle COB = \angle ACO + \angle CAO$

Hence:

\(\ds \angle ACO\) \(=\) \(\ds \angle CAO\) Isosceles Triangle has Two Equal Angles
\(\ds \leadsto \ \ \) \(\ds \angle COB\) \(=\) \(\ds 2 \angle ACO\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 2 \angle COB\) \(=\) \(\ds \angle ACO\)


Similarly, from External Angle of Triangle equals Sum of other Internal Angles:

$\angle COA = \angle BCO + \angle CBO$

Hence:

\(\ds \angle BCO\) \(=\) \(\ds \angle CBO\) Isosceles Triangle has Two Equal Angles
\(\ds \leadsto \ \ \) \(\ds \angle COA\) \(=\) \(\ds 2 \angle BCO\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 2 \angle COA\) \(=\) \(\ds \angle BCO\)


Therefore:

\(\ds \angle ACO + \angle BCO\) \(=\) \(\ds \frac 1 2 \angle COB + \frac 1 2 \angle COA\)
\(\ds \leadsto \ \ \) \(\ds \angle ACB\) \(=\) \(\ds \frac 1 2 \left({\angle COB + \angle COA}\right)\)

But $\angle COB + \angle COA$ equals two right angles.

Hence $\angle ACB$ is a right angle.

$\blacksquare$


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.


Sources

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