Thales

Thales of Miletus (624 – 546 BC)

Thales of Miletus was one of the seven sages of Greece and considered by Aristotle to be the first philosopher in the Greek tradition. The twentieth century philosopher Bertrand Russell goes further and says that Western philosophy begins with Thales.  As far as we are concerned, Thales was the first mathematician in that he saw the need for deductive reasoning. Others in the ancient world, the Babylonians and the Egyptians for example, certainly knew some of the gems of geometry and made good use of them in engineering and industry. It was Thales, however, who desired to prove these facts using deductive reasoning – starting with a set of axioms and making conclusions through deduction. Thales also wanted to understand the world not through mythology but through the human mind. He is often associated with the phrase “All is water.” To the modern ear, this sounds preposterous and overly simplistic. However, we can also view this expression to mean that the world can be understood by humans through a few basic principles. This was a enormous departure from the thinking of the ancient world. Unfortunately we can not read any of the original writings of Thales since they have been lost to time. What we do know about him and his works are through other Greek philosophers. Read more about Thales.

Below are four elementary facts we all learn in high school that are attributed to Thales – though there is some debate about which of these facts Thales actually proved in detail (see the Heath book, A History of Greek Mathematics, Vol I which is part of the assigned reading). These facts were known much earlier by others but Thales saw the need to put these facts on a firm foundation, i.e., prove them!

• A circle is divided by any diameter into two equal parts.

• The angles at the base of an isosceles triangle are equal.

• When two straight lies intersect, the opposite angles are equal.

• Two triangles are congruent if they have one side and two angles equal.

This following theorem on similar triangles one learns from high school is also attributed to Thales.

Theorem: Consider the following triangle ABC together with the line segment DE parallel to BC. Then |AB|/|AD| = |AC|/|AE|.

Proof: Link. I include the proof below.

Draw a line segment from B to E and another from C to E. Then draw a line segment from E to F that is perpendicular to AB and another line segment from D to G that is perpendicular to AC. See the drawing below.

Area(ADE) = 1/2 |AD| |EF| = 1/2 |AE| |DG| (base times height)

Area(BDE) = 1/2 |BD| |EF|

Area(CDE) = 1/2 |CE| DG|

Thus Area(ADE)/Area(BDE) = |AD|/|BD|  (1)

and Area(ADE)/Area(CDE) = |AE|/|CE|  (2)

Also note that

Area(BDE) = Area(CDE) (3)

since these two triangles share the same base DE and, since DE || BC they have the same height.

Combining equations (1), (2), (3) we see that

|AD|/|BD| = |AE|/|CE|

Taking inverses of both sides we get

|BD|/|AD| = |CE|/|AE|

and so, adding 1 to both sides we get

|BD|/|AD| + |AD/|AD| = |CE|/|AE| + |AE|/|AE|

which means that

(|BD| + |AD|)/|AD| = (|CE| + |AE|)/|AE|

But this means that

|AB|/|AD| = |AC|/|AE|

QED.

Here is another gem of Thales.

Theorem: If AC is a diameter of a circle and B is any other point on the circle (other than A or C), then the angle ABC is a right angle.

Proof: Consider the following drawing:

Since OB = OC (since they are all radii of the same circle), then the triangle OBC is an isosceles triangle and so, by one of Thales earlier results, the angles OBC and OCB are equal. Since OB = OA, then, again by Thales earlier theorem, the angles OAB and ABO are equal. Since the sum of the angles of a triangle, in particular the triangle ABC, must add up to 180 degrees it must be the case that a + (a + b) + b = 180. This means that 2 a + 2 b = 180, or equivalently a + b = 90. This shows that the angle at the vertex B is 90 degrees. QED.

Of course, missing from the proof of this theorem is a proof of the fact that the (interior) angles of any triangle sum of 180 degrees (or, in the language of Thales, “sum to two right angles”). Here is the proof of this gem: label the angles of your triangle A, B, C and draw parallel lines L and M in the image below

Now use some basic facts about angles (fill in the details) and you can see that A + B + C forms a 180 degree angle in the above drawing.

We should also point out that there is a converse to Thales’ theorem.

Converse to Thales’ theorem: A right triangles’ hypotenuse is a diameter of its circumcircle.

Proof: Take the right triangle and flip it across its diagonal to form a parallelogram. Note that the two diagonals will bisect each other. The point of intersection of these two diagonals will be a center of the circumcircle — with obviously the hypotenuse as a diagonal. QED.

Here is a nice corollary of this theorem which tells us how to construct a tangent to a circle – using only a straightedge and compass. We will talk much more about this in the angle bisection and angle trisection section later or. Right now just recall how to bisect an angle using a straightedge and compass. You might need to go over your high school notes.

Corollary: Given a circle C and a point P outside the circle, it is possible to construct, using only a straightedge and compass, a tangent to this circle which passes through P.

Proof:

Draw a line from the center O of the circle C to the point P. Now, using a straightedge and compass, draw the midpoint M of the line OP. Now draw a circle centered at M. Thus circle will intersect the original circle C at some point T. By Thales theorem (the one we just did), the triangle OTP has a right angle at T. By the definition of tangent, this says that the line PT is tangent to our original circle C. QED.

We’ll end our visit with Thales with a nice application attributed to him which finds the distance from a position on land to a ship at sea. Suppose we are on land at position A and a ship at out to sea at position B.

How do we find the distance from A to B. Here is an ingenious way to do it.

–Pick a point D (on shore) so that AD is perpendicular to AB. A basic straightedge and compass construction says we can do this.

–Pick a point E (again on shore) so that ED is perpendicular to AB. Again, a basic straightedge and compass construction says we can do this.

–Find the midpoint M of AD. Again, straightedge and compass.

–The line passing through M and B will pass through some point P on the line ED.

–Consider the triangles PDM and MBA. These triangles have right angles at D and A. By Thales, these triangles have equal angles at the vertex M. Since M is the midpoint of DA,  we have DM = MA. Thus, again by Thales, these two triangles are congruent.

–Thus the distance from A to B (the one we want to compute) is the distance from P to D (which is one we can measure).

Here is a drawing to make things clearer.