Thales' Theorem

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Thales' theorem:

If a triangle is inscribed inside a circle, where one side of the triangle is the diameter of the circle, then the angle opposite to that side is a right angle. The converse of this is also true. $_\square$

There are many ways to prove this theorem. One of the most classic proofs is as follows:

We know $AO=BO=CO$ as all of them are the radii of the circle. Hence, $\angle OAB=\angle OBA$ and $\angle OBC=\angle OCB$ because angles opposite to equal sides are equal.

Let $$\begin{aligned} \angle OAB&=\angle OBA=\alpha \\ \angle OBC&=\angle OCB=\beta. \end{aligned}$$ Then we have $$\begin{aligned} \angle ABC + \angle BCA + \angle CAB &=180^\circ \\ \angle OAB+(\angle OBA+\angle OBC)+\angle OCB&=180^\circ \\ \alpha+(\alpha+\beta)+\beta&=180^\circ \\ \alpha+\beta&=90^\circ \\ \angle ABC&= 90^\circ.\ _\square \end{aligned}$$

There's another way of proving this, which is based on another theorem called the alternate segment theorem which states that

Alternate segment theorem:

The angle subtended by a chord (or two radii) at the center of a circle is two times the angle subtended by it on the remaining part of the circle. $_\square$

Let us now try to prove Thales' theorem with the help of the above theorem.

According to the angle segment theorem, we have the following diagram:

$$\angle AOB = 2 \angle ADB.$$

Let us say that the radii $AO$ and $BO$ form the diameter, then the figure would look like this:

$$\angle AOB = 180^{\circ} \implies \angle ADB = \frac{180^{\circ}}{2} = 90^{\circ}.\ _\square$$

The longest chord in a circle is its diameter.

Let a circle with center $O$ have a chord $AB$. Join $OA$ and $OB$, and let $P$ be a point on $AB$ such that $OP\perp AB$.
We know that the perpendicular from the center of a circle to a chord bisects the chord. Then $PA=PB= \dfrac12 AB$.

Let the radius be $r$, the length of chord $AB$ be $c$, and the perpendicular distance between the center and the chord $AB$ be $x$.

Now, by the Pythagorean theorem, in $\triangle OPA$
$$\begin{aligned} OA^2 &=OP^2 +AP^2\\ r^2 &= x^2 + \left( \dfrac c2\right)^2 \\ \Rightarrow c &= 2\sqrt{r^2-x^2}. \end{aligned}$$ So, we need to find the maximum possible value of $c$, right? We know that a function is maximized when its derivative is equal to zero. Then, since radius $r$ is constant, $c$ is maximized when $$\begin{aligned} \dfrac{dc}{dx} &=& 0 \\ \dfrac d{dx} \Big (2\sqrt{r^2-x^2} \Big) &=& 0 \\ 2 \dfrac d{dx} \Big (\sqrt{r^2-x^2} \Big ) &=& 0 \\ 2 \left( \dfrac x{\sqrt{r^2-x^2}} \right) &=& 0 \\ \Rightarrow x &=& 0. \end{aligned}$$ Then, $c$ is maximized when $x=0$, implying that the maximum value of $c$ is $$c = 2\sqrt{r^2 - 0^2} = 2\sqrt{r^2} = 2r = 2 \times \text{(radius)} = \text{(diameter)}.\ _\square$$