Thales’ Theorem states the following:

Using the diameter of a circle as the base of a triangle with the apex on that circle, means that this triangle will be a right triangle and the diameter will be the hypotenuse of that triangle.

As John Page says, “The diameter of a circle subtends a right angle to any point on the circle”. (The word “subtends” means that it “creates an angle at a distant point”).

The diagram shows it more clearly. Position ‘O’ is the center of the circle. Line AB is the diameter of the circle. Point ‘c’ is on the circumference.

Thales’ Theorem says that the angle ACB of the triangle ACB is a right angle…regardless of where ‘C’ may be placed on the circumference.

## To Prove Thales’ Theorem

To prove Thales’ Theorem, let’s start by adding another line, CO, to the diagram.

## Preparing for the Proof of Thales’ Theorem

The point ‘O’ is the center of a circle with radius of length ‘r’. AB is a diameter with ‘O’ at the center, so length(AO) = length(OB) = r. Point C is the third point on the circumference. We have triangles OCA and OCB, and length(OC) = r also.

Triangle OCA is isosceles since length(AO) = length(CO) = r. Therefore angle(OAC) = angle(OCA); let’s call it ‘**α**‘ (“alpha”). Likewise, triangle OCB is isosceles since length(BO) = length(CO) = r. Therefore angle(OBC) = angle(OCB); let’s call it ‘**β**‘ (“beta”).

(If anyone asks, we can get into a discussion of this property of isosceles triangles at another time. That’s the purpose of the “comments” section, in my humble opinion.)

Since AOB is a straight line, angle (AOB) = 180. So angle(AOC) + angle(COB) = 180. Let’s record this fact as angle(COB) = 180 – angle(AOC).

The sum of the interior angles of a triangle is 180. So in triangle OCA, the sum of angles OAC + OCA + AOC = 180. As we said before, this is **α**+**α** + angle(AOC) = 180, or 2***α** + angle(AOC) = 180.

Likewise, in triangle OCB, angle(OCB) + angle(OBC) + angle(COB) = 180. As noted, we can say **β**+**β** + angle(COB) = 180, or 2***β** + angle(COB) = 180.

## First Proof of Thales’ Theorem

We also see that triangle ACB is a triangle, so its interior angles add up to 180 also. So angle(BAC) + angle(ABC) + angle(ACB) = 180. That was **α**+**β** + angle(ACB) = 180.

But clearly angle(ACB) is the sum of angle(ACO) + angle(OCB), which are values **α** and **β**. So **α**+**β** + **α**+**β** = 180. So 2*(**α**+**β**) = 180. Dividing both sides by 2 shows that **α**+**β** = 90. This is precisely what Thales’ Theorem says, so it is proven.

## Second Proof of Thales’ Theorem

The first proof used the fact that angle(OAC) is precisely angle(BAC). The viewpoint changed from the two smaller triangles to the one triangle made up of the two smaller ones. This proof will take a different viewpoint.

First, let’s remember that angle(COB) = 180 – angle(AOC). So the previous sum of triangle OBC’s angles could be written as 2***β** + ( 180 – angle(AOC) ) = 180.

We have two equations: 2***α** + angle(AOC) = 180, and also 2***β** + ( 180 – angle(AOC) ) = 180.

For this alternate proof, let’s add the angles of the two smaller triangles. Therefore ( 2***α** + angle(AOC) ) + ( 2***β** + ( 180 – angle(AOC) ) ) = 360.

Rearrange the terms and we find ( 2***α** + 2***β** ) + angle(AOC) – angle(AOC) + 180 = 360.

The beautiful part of this equation is that it explicitly shows that angle AOC does not matter: it is added and subtracted, thus cancelling itself out. The first proof simply stopped talking about angle AOC.

Bring out the common factor of “2”, cancel the “angle(AOC)” terms, and subtract 180 from both sides to show 2*(**α** + **β**) = 180. Divide both sides by 2 and we have returned to Thales’ Theorem, because **α** + **β** = 90.

## One Practical Application of Thales’ Theorem

A simple but practical application of Thales’ Theorem is to **find** the center of a circle, assuming you can draw a couple of right-angle triangles over it.

Draw a circle. Choose 1 point, “A”, on the circumference. Draw a right-angle triangle with that point at the right angle, making the triangle overlap the circle and extend beyond it.

Mark the places where the smaller sides (not the hypotenuse) cross the circle. These are points “B” and “C”. The line BC is a diameter, by Thales’ Theorem.

Choose a different point, “D” on the circumference, and draw a new right-angle triangle DEF. Once again, line EF will be a diameter. Their intersection is point “O”, the center of the circle.

## Who Was Thales?

Thales was a Greek philosopher. Based in Miletus, he studied geometry in Egypt. Several stories have survived about this early Greek philosopher, geometer and astronomer.

Two stories of astronomy and geometry seem quite credible. Thales predicted an eclipse in 585 BCE. He could determine how far away a ship was sailing by taking observations from two land positions.

The story that may reflect luck more than science involved cornering the market on olive press capacity. He overcame the stigma of poverty by predicting a bumper crop in olives, then buying up the rights to use olive presses for the upcoming harvest season. He then charged heavily for these presses.

Thales’ major claim to fame, however, is Thales’ theorem as highlighted in this article.

**References**:

Thomas Knierim, editor, The Big View, “Thales“, accessed on June 10, 2011.

Jim Loy, “Isosceles Triangles“, 1999, accessed on June 10, 2011.

DB Education Services Ltd, “Thales Theorem – Proof of the Thales-theorem“, 2000-2011, accessed on June 10, 2011.

John Page, Math Open Ref, “Thales’ Theorem“, 2009, raccessed on June 10, 2011.

John Page, Math Open Ref, “Finding the center of a circle using any right-angled object “, 2009, accessed on June 10, 2011.

[…] of prior mathematicians such as Thales and Pythagoras. (Thales is best known for his theorem about circled triangles, and everyone knows the Pythagorean Theorem about the squares adjacent to right-angle […]