Thales' Theorem
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Theorem
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. \(_\square\)
Proof
There are many ways to prove this theorem. One of the most classic proofs is as follows:
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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 \[\begin{align} \angle OAB&=\angle OBA=\alpha \\ \angle OBC&=\angle OCB=\beta. \end{align}\] Then we have \[\begin{align} \angle ABC + \angle BCA + \angle CAB &=180^\circ \\ \angle OAB+(\angle OBA+\angle OBC)+\angle OCB&=180^\circ \\ \alpha+(\alpha+\beta)+\beta&=180^\circ \\ \alpha+\beta&=90^\circ \\ \angle ABC&= 90^\circ.\ _\square \end{align}\]
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:
\[ \angle AOB = 180^{\circ} \implies \angle ADB = \frac{180^{\circ}}{2} = 90^{\circ}.\ _\square \]
\(AB\) is the diameter of a semi-circle.
\(C\) is a point on the circumference.
What is the measure of angle \(ACB?\)
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 \(x\).
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Now, by the Pythagorean theorem, in \(\triangle OPA \)
\[\begin{align} OA^2 &=OP^2 +AP^2\\ r^2 &= x^2 + \left( \dfrac c2\right)^2 \\ \Rightarrow c &= 2\sqrt{r^2-x^2}. \end{align} \] 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 \[ \begin{eqnarray} \dfrac{dc}{dx} &=& 0 \\ \dfrac d{dx} \Big (2\sqrt{r^2-x^2} \Big) &=& 0 \\ 2 \dfrac d{dx} \Big (\sqrt{r^2-x^2} \Big ) &=& 0 \\ 2 \left( \dfrac x{\sqrt{r^2-x^2}} \right) &=& 0 \\ \Rightarrow x &=& 0. \end{eqnarray} \] Then, \(c\) is maximized when \( x=0\), implying that the maximum value of \(c\) is \[ c = 2\sqrt{r^2 - 0^2} = 2\sqrt{r^2} = 2r = 2 \times \text{(radius)} = \text{(diameter)}.\ _\square\]
True or false?
The length of the largest chord possible in a circle is always equal to twice its radius.