This web page is reproduced from the author’s book *Principles of nature; towards a new visual language, WA Roberts P/L Canberra. 2003.* [Used with permission of the author] Edited and reformatted for the web, including animated diagrams and additional commentary by the author.

**Introduction**

Scale structure theory has implications not only to the units and measurement of areas but also to triangle classification as we have seen. Particularly pertinent to the theorem presented here on this page is the implementation of the equitriangular unit of area (etu) as we defined earlier. We will also later use our new knowledge of the area of a eutrigon (in terms of these relative units of area, *etu*) in the *algebraic interpretation* of the geometric construction (figure *ET2*) below — a form of *resonant scale structure* (in the terminology of the new theory) — that *visually proves the theorem*.

### the Eutrigon Theorem

**—Geometric Form**

The area of any eutrigon (shaded in figure *ET1*) is equal to the sum of the areas of the equilateral triangles on its legs *a* and *b*, minus the area of the equilateral triangle on its hypotenuse, *c*.

Geometric form of Eutrigon TheoremThe area of any eutrigon is equal to the sum of the areas of the equilateral triangles on its legs

aandb, minus the area of the equilateral triangle on its hypotenuse,c.(W. Roberts, 2003, p.122)

*A look-and-see proof*

**Note: all triangles which ***appear* equilateral *are* equilateral.

### Detailing the proof (see figure *ET3*)

From Figure *ET3* it can be seen that:

1) Triangles marked “Q” are *congruent eutrigons* since each contains one 60° angle, and their respective sides are equal.

2) Triangles A, B, and C are equilateral

3) A and B are the equilateral triangles on the legs of eutrigon Q, and C is the equilateral triangle on its hypotenuse. >

4) The overall diamond shape is a rhombus consisting of an upper and lower equilateral triangle of *identical area*.

5) Point “4” means that, expressed in terms of areas,

A + B + 2Q = C + 3Q,

and thus, Q = A + B – C

therefore the ** Eutrigon Theorem** is proved.

**Figure ET3**

Now, some readers may not be satisfied that in the ‘look & see’ proof given above, that eutrigon Q may be *any* eutrigon . Before we prove this more rigorously let us *imagine* whether it is so.

Imagine an *animation* of equilateral-triangle C rotating while maintaining its equilateral shape but varying in size such that its vertices ‘slide along’ the outer equilateral triangle it is touching…[I’ve provided such an animated version below and added some colour to jazz it up a bit! It auto-loops a few times with a pause in between. If it has already been through its routine by the time you get to this part of the page, just click the ‘refresh button’ in your browser window, and it will start over.]

Figure *ET4
*(animated, © 2003 copyright W Roberts. All rights reserved)

For an elaboration of the proof that “Q” in Figure *ET3 *covers every possible eutrigon shape.

See here for PDF [173kb] of geometric form of Eutrigon Theorem from the book (W. Roberts, *Principles of Nature: towards a new visual language*, 2003, pp. 122-125)