Thales of Miletus

Thales of Miletus

Thales of Miletus

Introduction:

[pic 1]

Thales of Miletus was born around 625 BC and died 78 years later. He was a Greek astronomer, mathematician and philosopher according to some historians, he would have been a political figure. I lived in Miletus, in actual Turkey.

t is difficult to know exactly what his achievements in mathematics and geometry are because there are no writings or sources from this period and some stories and anecdotes about him and his life might be exaggerated.

The following discoveries are most of the time credited to him: the intercept theorem (which will be addressed in this file), and few others about triangles including one about angles in a circumscribed triangle (which will also be addressed in this file).

[pic 2]

Theorems and demonstrations

The Thales theorem[pic 3]

Theorem

If one side of a triangle is the diameter of its circumscribed circle hence this triangle is rectangle. The circumscribed circle of a triangle is the circle that goes through every vertex.

Demonstration

This theorem can be proved using angles and equilateral triangles. For this demonstration, we need to pre-establish that the sum of the angles of a triangle is equal to 180° and that in an isosceles triangle two angles are equal. We take a circle of centre A and a diameter [BC]. We add a point D on the circle. We draw the segment [AD], it gives us two triangles ∆ABD and ∆ACD. Considering ∆BCD angles we can write that:[pic 4]

The sum of those angles can be written as:

As B, C and D are on the circle AB=AC=AD. So ∆ABD and ∆ACD are two isosceles triangles which means that angles α and α’ are equals and ε and ε’ too.

It proves that in every case, if in a triangle as one of its sides [AB] as the diameter of a circle and that the third point C is on the circle, then the angle ∠ACB equals 90°.

The intercept Theorem

Theorem

[pic 5]

O is the intersection of (AC) and (BD).

(CD) // (AB)

So according to the intercept theorem

[pic 6]=[pic 7]=[pic 8]

Demonstration[pic 9]

1) Equal area of the triangle

ABCD is a rectangle, so (AB)//(DC) and AD=BC. F and E belong to (AB).

The area of a triangle is given by the relation A= [pic 10]

Considering the triangles DFC and DCE we can say that they have the same base [DC], and same height [AD] or [BC].

In the figure above, the area of the triangle DFC in blue is:

 Adfc=[pic 11]= (DC*BC)/2

The result is the same for the triangle DCE in red:

Adce =[pic 12]= (DC*BC)/2

We can conclude that:

Adfc = Adce

So, we can say that if two triangles have their third top on a parallel line of the common base, then the areas of the two triangles are equal.

[pic 13]

In this figure, ABO is a nondescript triangle, but (AB) // (CD)

We see that for ACD in red and CDB in green, these two triangles have a common base [CD] and their third tops are on (AB) which is parallel to (CD).

So Aacd = Acdb

Now considering triangles ODA and OCB: they are equal to the sum of the area of OCD plus respectively the area of ADC and the area of BCD.

But we know that the areas of BCD and ADC are equal, so the areas of ODA and OCB are equal

So, we can write Aocd / Aoda =Aocd / Aocb

If we substitute it by the relation of the area of a triangle and write:

Aocd= (OC*h1)/2 = (OD*h2)/2

[pic 14]

  [pic 15][pic 16][pic 17][pic 18]

by simplifying by [pic 19] on the left side and by[pic 20]on the right side.

We find [pic 21]=[pic 22] which gives the first part of the theorem.

2) Equal ratio

If a/b= c/d with b and d different of 0

Then by multiplying by d and b:

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