Thales' Theorem

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Thales' Theorem is a simple yet important geometrical result which describes the relationship between inscribed right angles and circles. A special case of the inscribed angle theorem, Thales' Theorem is as follows:

Let $A$ and $C$ be two points on the circumference of some circle $S$ such that $\overline{AC}$ is a diameter of $S$. If a point $B$ is placed anywhere on the circumference of $S$ (other than $A$ or $C$), then $\angle ABC$ will always be a right angle.

Using the inscribed angle theorem, it also follows that the reverse of this statement holds. Given any 3 points $A,B,C$ on the circumference of $S$ such that $\angle ABC=90^{\circ}$, $\overline{AC}$ must be a diameter of the circle. We now present two different approaches to proving Thales'.

A proof via inscribed angle theorem.

Consider a circle $S$ with center $O$ and two diametrically opposite points $A,C$ on the circumference of$S$. Since points $A,C,O$ are all co-linear, it follows that $\angle AOC=180^{\circ}$.

By the inscribed angle theorem, we know that any angle $\theta$ inscribed in $S$ will be half of the central angle which subtends the same arc on $S$. In other words, if a point $B$ is placed on the circumference of $S$, then angle $\angle ABC$ will be half of angle $\angle AOC$. Since we have that $\angle AOC=180^{\circ}$, it follows that $\angle ABC=90^{\circ}$

Thales' theorem has many elegant applications to pure geometry. However, since Thales' often arises in coordinate geometry, let's take a look at a proof of Thales' which uses just that.

A proof via coordinate geometry.

Consider a circle $S$ of radius $r$ located at the origin. Notice that $S$ is described by the equation $x^2+y^2=r^2$. Consider the two diametrically opposite points $(-r,0)$ and $(r,0)$ located at the intersection of $S$ with the $x$-axis. Place a third point anywhere on the upper half of the circumference of $S$.

Pythagoras' theorem guarantees that in a triangle with side lengths $a,b,c$, with $c$ being the longest side, $a^2+b^2=c^2$ if and only if the triangle is a right triangle. Therefore, we will show that $\theta=90^{\circ}$ in the diagram to the right by showing that $a^2+b^2=c^2$.

Using Pythagoras' theorem, we can determine the legnths of each side as follows

$\begin{aligned} a^2&=(x-(-r))^2+(y-0)^2=x^2+2xr+r^2+y^2\\b^2&=(x-r)^2+(y-0)^2=x^2-2xr+r^2+y^2\\c^2&=(r-(-r))^2=4r^2\end{aligned}$

Next, observe that

$a^2+b^2=(x^2+2xr+r^2+y^2)+(x^2-2xr+r^2+y^2)=2x^2+2y^2+2r^2=2(x^2+y^2)+2r^2$

Recall that since the point $(x,y)$ was chosen on the circumference of $S$, it must satisfy $x^2+y^2=r^2$. Using this to simplify the equation above:

$a^2+b^2=2(x^2+y^2)+2r^2=2(r^2)+2r^2=4r^2=c^2$

Since we have shown that $a^2+b^2=c^2$, we have also proved that $\theta=90^{\circ}$ which finishes our proof of Thales'.

There are many other ways to prove Thales', some using pure geometry, some using trigonometry, and some even using complex numbers! We encourage you to explore and try to come up with your own unique proof of Thales'. If you do manage to come up with your own proof, great! Just don't get too excited and sacrifice an ox (see entry "Thales' theorems" on this page for more details).

Here are a couple problems centered around Thales' for practice. The first few aren't too bad, but be careful, the last one is quite tricky!

Circle $S$ is placed in the 2D-plane with radius $3$ and its center $O$ placed on the $y$-axis. $S$ intersects the $y$-axis at both the point $G$ and the origin (labeled $F$).

Circle $T$ is then drawn with twice the radius of circle $S$ such that $T$ also intersects $F$ and $G$.

If circle $T$ intersects the $x$-axis at points $X$ and $G$, find the length of segment $\overline{FX}$.

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Solution

Since the $x$-axis and $y$-axis are perpendicular, $\angle XFG=90^{\circ}$. Combined with the fact that $\Delta XFG$ is inscribed in $T$, Thales' theorem tells us that $\overline{XG}$ is a diameter of $T$.

The problem is finished by using Pythagorean theorem on $\Delta XFG$ to get $|\overline{XF}|=\sqrt{12^2-6^2}=\sqrt{108}$.

Circles $S$ and $T$ are co-centric at point $O$ with circle $S$ inside circle $T$. Line segment $\overline{BA}$ is a cord of circle $T$ and is tangent to circle $S$ at point $D$. Line segment $\overline{BC}$ is a cord of circle $T$ and is perpendicular to segment $BA$.

If $\overline{BC}$ has length 1 and $\overline{BA}$ has length 2, find the radius of the inner circle $S$

Hint

Use the Pythagorean theorem on $\Delta OAD$ to find $\overline{OD}$ once you've found the length of $\overline{OA}$.

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Solution

By Pythagorean Theorem, $|\overline{AC}|=\sqrt{5}$. Since $\angle ABC=90^{\circ}$, Thales' guarantees that $\overline{AC}$ is a diameter of the outer circle. Since both circles share the same center $O$, we know that $\overline{AC}$ passes through the center of both circles. Therefore, $|\overline{OA}|=\frac{\sqrt{5}}{2}$ since it is a radius of the outer circle.

By symmetry, we know that $D$ is the midpoint of $\overline{AB}$ and therefore $|\overline{AD}|=1$.

Finally, using Pythagorean theorem on $\Delta OAD$, we get $|\overline{OD}|=\frac{1}{2}$.

This last one can be a bit tricky.

Circle $T$ is drawn with perpendicular cords $\overline{AB}$ and $\overline{BC}$ with lengths 3 and 4, respectively. Circle $S$ is then drawn such that $\overline{AB}$ is a diameter of $S$. Finally, a line is drawn through point $A$ perpendicular to $\overline{AC}$. Let point $E$ be the other intersection of this line with circle $S$.

Find the length of segment $\overline{BE}$.

Hint 1 (view this before Hint 2)

Draw a perpendicular line from point $B$ to $\overline{AC}$ and label their intersection $X$. Notice that $BX$ is parallel to $\overline{AE}$. Do $\overline{BX}$ and $\overline{EA}$ have the same length? (Use Thales!)

Hint 2

Consider using segments $\overline{BX}$ and $\overline{AC}$ to calculate the area of $\Delta ABC$. How else can we determine the area of $\Delta ABC$?

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Solution

Draw a perpendicular line from point $B$ to $\overline{AC}$ and label their intersection $X$. Notice that $BX$ is parallel to $\overline{AE}$. By Thales', we know that $\angle AEB=90^{\circ}$ so $AEBX$ is actually a rectangle and therefore $|\overline{AE}|=|\overline{BX}|$ and $|\overline{EB}|=|\overline{AX}|$.

By pythagorean theorem, we have $\overline{AC}=5$. Since segments $\overline{AB}$ and $\overline{BC}$ are perpendicular, the area of triangle $\Delta ABC$ is 6. We can also find the area of triangle $\Delta ABC$ by using segments $\overline{BX}$ and $\overline{AC}$, specifically. $\begin{aligned} \text{area}(\Delta ABC)&=6\\ \frac{1}{2}|\overline{BX}||\overline{AC}|&=6\\ \frac{1}{2}|\overline{BX}|5&=6\\ |\overline{BX}|5&=\frac{12}{5}\end{aligned}$

Since $\overline{AE}=\overline{BX}=\frac{12}{5}$, it follows from Pythagorean theorem that $\boxed{|\overline{EB}|=\frac{9}{5}}$