Addendum 3.1.1 to

3.1 Analogical Thinking

 

 

Analogical thinking in Pythagoras’s theorem

 

Levels of reality

Pythagoras’s theorem can be proven along multiple routes. For our exploration on this website, the evidence offered by analogical thinking is of special interest. It enables us, through a common factor as intermediary, to represent the correlations in the realm of analogy as a (mathematical) law in the physical realm.
This conversion from the analogical to the physical level hinges on this intrinsic intermediary, the so-called arithmetic mean. Because of the congruence between the original triangle and the two smaller derivative triangles within it, all line segments fill different, alternating roles in relation to each other. This means that the analogical correlation can now also be expressed in terms of an absolute value.

 

However, the arithmetic mean does not appear in the final formulation of the theorem. The dimension in which it functions has been lost by the projection of the analogical correlation into a (mathematical) law pertaining to a flat surface. We might say that the arithmetic mean serves as a transferring agent from one level of reality to another.

 

An example

By way of example, we take the right-angled triangle ABC, with its 90° angle in A and its three sides a, b and c.
By dropping a perpendicular line from A to c, we get three identical triangles: ABC, ADC and BDA.


driehoek

fig 3.1.1 The Pythagorean triangle

In those three triangles, the functions of the line segments change places: The length (a) of the big triangle (ABC) is also the hypotenuse of the middle-sized triangle (ADC).
The same goes for the short side (b) of the original triangle (ABC): in triangle ADB, b has become the hypotenuse.
So, because of the analogical correlation, if we know the lengths of the line segments a and b, we can also deduce the lengths of the line segments d and c-d (and therefore c).

 

Intermediary

Providing proof for this correlation does not fall within the scope of this website, but I do want to point out the essential roles of both the line segment AD, as the arithmetic mean, and point D on the line BC. In order to further clarify this, I have included the derivation of the formula below, at the bottom of this page.
For our line of reasoning, the main point is even though the arithmetic mean is of essential importance for the realization of the role reversals in the analogous relations between the three line segments, yet, at the same time, the arithmetic mean does not show up in the final formula, where it plays no role. It would seem that the arithmetic mean only serves as a ‘hinge’ between one realm of reality and the other. Belonging to the realm of analogous correlations, in the final formula it has stepped backed into the wings, together with the internal analogous correlation in question.

 

Direct relationship

If we now return to the original triangle ABC, it turns out, based on analogical reasoning, that the arithmetic mean enables us to formulate the direct relationship between the three line segments a, b and c, in the form of a ² + b ² = c ² .

 

Potential power

So the proven analogical relationships within any right-angled triangle can be transferred to the realm of physical reality, where it reveals itself as a (mathematical) law that applies to all right-angled triangles.

 

At a right angle, the analogical correlation has the potential to show up as a law in the realm that is perpendicular to it, such as the realm of physical appearances.

For our insight in astrology, this potential of the analogical correlation may well turn out to be crucial.

 

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Extra addendum to Pythagoras’s theorem

 

Derivation (fig.3.1.1)

For the right-angled triangles ABC and ADC, the following applies:
1. For the hypotenuse and the long leg:
b : d = c : b
After multiplication, this becomes b ² = d x c .
2. For the hypotenuse and the short leg:
a : (c - d) = c : a
After multiplication, this becomes a ² = (c-d) x c .

By adding up both equations, we get:
a ² + b ² = d x BC + (c-d) x BC
a ² + b ² = (d + (c-d)) x BC
a ² + b ² = c x c
a ² + b ² = c ²
Quod erat demonstrandum.

This concludes this extra addendum.

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