## Basic „A“ Trisection of an Angle

**FIRST BASIC TRISECTION OF AN ANGLE
SCHEME „A“**

(RELEVANT FOR ANGLES UP TO 90° WITH EXAMPLE OF

RANDOMLY GIVEN ANGLE)

**Basic „A“ Trisection of an Angle with an example**

In this chapter we will demonstrate by pure geometry, but in keeping with the rule (only using a compass and ruler without markings) the basic trisecting of a hexagonal segment by putting it under the mark of “A”. The reason for this is quite plain. To comprehend the system of trisecting and its product is not only to get three equal and overlapping isosceles triangles, but also several other products of trisecting so as to render a clear depiction of trisecting in light of our first encounter with it. As for some other products, we will “geometrically” leave them out. Namely, let us repeat: One of the basic laws of ancient, sacred or natural geometry is that all divisions of circles of whatever radius are drawn as full circles (full angles) regardless of how many there are in any division, with no markings – whether by letters or numerals. Why?

By not drawing like that we omitted an array of particulars. In the second place, if one observes the “A” scheme of this chapter, how would any beginner be able to comprehend it, especially if distinguishing marks of numerals or letters were added; something the current “incomplete” geometry does.

In addition to knowledge, ancient geometry calls for exceptional preciseness since every imprecise arc minute leads to an imprecise result. We have selected trisecting as our initial “A” scheme for another reason as well: besides trisecting, to do away with one more geometric inaccuracy that is attendant in present-day teaching of geometry, the roots of which go back to Archimedes.

This will be discussed in one of our subsequent chapters.

Let us recapitulate: It is essential to remember scheme “A” trisection and the example of trisecting its randomly given angle.

* * *

In ancient geometry the hexagon is the basis of all geometric beginnings, namely, the division of the central circle with its radius into six equal parts. This flower-like design is the base of whatever ancient geometrical construction, be it an angle, a geometrical figure or a geometrical body, their divisions and their products. It is good to remember: the basis – the beginning.

* * *

It is acknowledged that one part of the hexagon is an equilateral triangle, and its chord is equal to the radius of the basic circle. However, the aim of the trisecting procedure is not to divide its chord, but its arc. It is therefore necessary to focus on the chord’s 1/6 of the hexagon, while its sides only serve as the edges of the 60° angle. 360°÷6 = 60°

* * *

Since we have properly drawn the hexagon (the basic circle with its six dissevered circles) we can immediately divide the chord of one component into two equal component parts (intersection – center of central circle – intersection of two dissevered circles). This means: if we connect the intersections of the dissevered circles with the center of the basic circle, we will get its division into 12 equal parts, in other words, a dodecagon! But, let’s keep focused only on the one-sixth.

* * *

Therefore, our goal is to divide 1/6 of the hexagon into 3 equal parts. Now the intersection of the chord and bisector become the center of a new, smaller circle whose radius is ½ of the chord of one hexagonal segment (for clarity of the drawing it is depicted in another color, although in the geometry of ancient times markings were not indicated in letters nor in numerals.)

* * *

The new circle divided into six parts, however, does not divide the arc of the basic circle, namely, of its hexagonal segment into three components even though it is essential for the division of the arc of the hexagonal segment. We therefore say that this is the first phase of the trisection segmentation. For an easier percept, we will not geometrically draw this circle correctly with its dissevered circles, since we only want to find the division of the arc on one segment of the basic hexagon.

* * *

Once again the bisector divides the small circle into two parts, meaning: if the peaks of the chord of the basic hexagonal segment are connected with the bisector’s intersection of the small circle, we get a right angle (90°).

* * *

The 90° angle arc – right angle; even though it is on the same bisector and has common peaks with the peaks of the hexagonal chord peaks of the basic circle and the small circle’s poles, it is larger than the arc of the hexagonal segment of the basic circle.

* * *

Its arc, the hexagon partitioning of its small circle, dissevers it into three equal parts, but not the arc of the hexagonal segment of the basic circle; and once again, it is essential for partitioning of the arc of the hexagon segment – which is our aim. (For the purpose of distinguishing the phases, we apply another color.) This is, therefore, the second phase of dividing the hexagon segment of the basic circle into three equal parts.

* * *

As can be seen, the second phase implies the partition of the 90° arc of the small circle into three equal parts. Had we now connected the intersections of the 60° – 90° arcs with the center of the basic circle, the trisecting would be faulty even though they all have the same bisector and poles as well (poles of the hexagon segment of the hexagon’s basic circle.) Their arcs are of different sizes solely on the partitioned poles of the 90° arc of the small circle (the second phase of partition) towards the center of the basic circle of the arc’s 1/6 segment, which implies that the “points of trisecting” come forth on its arc. That is then the trisection of the 60° angle of one segment of the basic circle.

* * *

That is how we come to the correct trisecting of the 60° angle: three isosceles triangles that are equal, namely 60°÷3 = 20°. This way or scheme is used for all angles up to 120°. How? First of all we will show an example of dividing a randomly given angle up to 90°.

* * * * * *

**Example of trisecting randomly given angle
**

**up to 90° according to trisecting scheme „A“**

Draw the randomly given angle and define its arc at random: inscribe the chord of its arc and bisect it into two equal parts.

* * *

The chord’s center of a randomly given angle is the center of the segment of the circle and the chord’s end points take over the role of poles (subtended) of the smaller circle.

* * *

Having defined the poles of the circle’s chord, we divide it (concisely, by omitting the six divided circles) into 6 parts, i.e. with the radius of ½ chord of the randomly given angle and the angle’s arc.

* * *

In this way we have reached the first phase of trisecting according to scheme “A” trisection of angles up to 90°.

* * *

The bisector of the given angle also divides the chord’s circle into two parts. If we connect the circle’s half-way point with the chord’s end points of the given angle, which are simultaneously the subtended poles of the circle’s ½ chord of the given angle, we get an angle of 90° – a right angle, whose arms pass thru the ends of the chord.

* * *

The 90° angle – the right angle, has an arc that differs from the given angle’s arc – (it is larger) and from the arc of the chord’s circle (it is smaller) but it has common poles, i.e. peaks of the randomly given angle’s chord.

* * *

Rays from the circle’s chord, and from its subtended hexagonal poles towards the 90° vertex divide the angle into three equal sections.

* * *

Thus we have finished the second preparatory phase in accordance with trisecting scheme „A“. As we shall see, it is the “essence” – one of the elementary patterns by which trisecting is conducted.

* * *

Then, and only then, from points of the tri-sectioned division of the 90° angle, chords, circles of ½ chords, do the rays extend toward the peak of the given angle and divide the angle’s arc into three equal parts; or to put it more precisely, from the points of dividing the arc of the 90° angle towards the vertex of the randomly given angle.

* * *

We thereby achieved the trisecting or dividing of the angle’s arc into three equal sections, and if we can dissever any angle into three equal sections, then it is clear that we have three equilateral triangles (apart from the hexagon). This is a pure trisection of an angle with the use of an unmarked ruler and compass, so simple, even though it is one of the more demanding (there are simpler ways)! Then why have we taken this example to begin with? This will be explained in one of our subsequent chapters.

* * * * * *

These are impressive circle drawings. To be honest, I got lost somewhere at picture 209. How do you get the three isosceles triangles of picture 210? To get some understanding of your method, it would be helpful to highlight how Archimedes’ trisection emerges in your drawings.

Poštovani gosp. Tomo,

Svaka Vam čast na znanju, trudu i upornosti! Ima nas možda malo koji sa velikim zanimanjem čitaju i proučavaju ova znanja ali nas ima. Možda je Vašim bližnjima teško razumjeti Vaša nastojanja kao što je i našima teško shvatit što mi u tome vidimo. Mi vidimo važne i velike istine od kojih je sazdan svijet i univerzum. Vaš rad vidim kao očuvanje velikih blaga, nadu da će se čovjek vratiti na put istine i ljepote. Beskrajno Vam hvala! Svako dobro Vam želim!

B.