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Dual angle trisection according to scheme „B“

DUAL ANGLE TRISECTION ACCORDING TO SCHEME „B“
FOR ANGLES LARGER THAN 90° WITH EXAMPLE

Just like in the first part of trisection scheme „B“, trisecting an arbitrary angle, in this second part we repeat the process of trisecting, but with a small divergence: instead of one segment, we trisect two segments as if each one is simultaneously separate; which is consistent with the comment in the first part of the preceding chapter that it is possible to trisect any angle larger than 90°, and not only that, but we likewise forecast the logical conclusion: if any angle in excess of 90° can be trisected, then that really means all angles. But let’s go step by step.

* * *

We will continuously encounter the groundwork – the hexagon: division of the circle with its radius; in other words, the partitioning of its own arc.

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We highlight two segments of the hexagon whose joint vertex is the center of the initial circle.

* * *

Each segment has its chord and its bisector, and their common point is the center of the initial circle and the chord’s vertex; the other pole of the hexagon.

* * *

Each segment has its circle whose centers are the intersections of the bisectors and chords, and their diameter is that of the chord.

* * *

Link the three circles from endpoints of the chord of same diameter (chord diameter of the segments) to the two circles of the hexagon segment.

* * *

Thereafter link four more circles from the centerline intersections of the segment circles (like in the first part of scheme “B” trisection of the hexagon segment – but doubled). Now we have a dual four-leaf flower-like pattern.

* * *

Each circle of each segment of the initial circle’s hexagon has its own right angle.

* * *

Now each right angle has its own arc which is larger than the arc of its hexagon segment, despite the fact of having common vertexes (poles) with the arc of the hexagon segment of the initial circle.

* * *

We mentioned in the first part of scheme „B“ that the lengths connecting the vertices of the four-leaf pattern are actually diagonals of the descriptive square of the hexagon segment’s circle based on the chord diameter and they divide the arc of the 90° angle into three equal parts.

* * *

And so we acquire the points of the second phase of trisection that are the grounds for trisection of the hexagon segment (this time doubled – two segments respectively, each with their own points of trisection).

* * *

Let’s reiterate: As ascertained, we can divide the arc (of each segment) into three equal parts only from these points towards the center of the initial circle.

* * *

In this way we have acquired three or respectively six isosceles triangles whose complementary angles (common vertices of the initial circle) are 20°.

* * *

But we are not interested in a segment by segment approach, as we were in the first part of trisection scheme “B”, but instead our interest is in two hexagon segments: 60°+60°=120°÷3=40° so we therefore take every other division of the arc of two hexagon segments. That is the version “B” scheme for trisection of angles in excess of 90°.

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EXAMPLE OF DUAL ANGLE TRISECTION ACCORDING TO SCHEME „B“
FOR ANGLES LARGER THAN 90° WITH EXAMPLE

On observation of these trisection „tools“, and being knowledgeable about the complete and full trisection arrangement and how it occurs, something in me nevertheless “broke”:   particularly in light of the beauty of the fully plotted schemes “A” and “B”– their genetics, their compatibilities, their productions. Many would be divested of the beauty of a perfect harmony. Therefore, instructed by my spirit, I realize that there are still many more themes worthy of being represented. There are other enigmas and allegedly other “impossibilities”; other never recorded (geometrically drawn) novelties in the written or depicted history of the world. But …

In the following chapters it is precisely the above mentioned “but” that impels me to present the two “genes” of trisection, each separately, in their full splendor only by using a compass and straightedge By way of trisection their compatibility becomes fully apparent, as well as the possibility to construct the nonagon and the 18-sided polygon. However, this is a field as extensive as trisection. Therefore it deserves six or seven or more chapters in some other book. So let’s proceed gradually. Let’s first of all focus on the example of trisection of the angle larger than 90°.

* * *

Thus our arbitrary angle is larger than the right angle with its randomly determined arc and bisector.

* * *

Each part of the arc (angle) divided by the bisector has its own chord.

* * *

Once again we divide each chord with bisectors.

* * *

Each chord cut with a bisector is the center of each circle of the chord’s diameter.

* * *

And from the vertices (acmes) of the chords we circumscribe circles of same diameter (that is to say, three).

* * *

Each bisector bisects the chord’s diameter into two parts. From bisects of the arcs and bisectors we circumscribe the circles of same chord diameter, that is to say 2+2=4.

* * *

Each of the inscribed circles has a right angle.

* * *

Each one of them has an arc that is larger than the arc of the arbitrary angle even though their segmented poles are the same.

* * *

We have acquired two four-leafed flower-like patterns. And we have said that the lengths linking the subtended vertices of the four-leafed patterns are diagonals of the descriptive square of the inscribed circles.

* * *

They divide the arc of their right angles into three equal parts.

* * *

So, we have acquired reduplicated trisection points from which half-lines extend towards the center (vertices of arbitrary angle) to divide the arc of the arbitrary angle into three equal parts twice.

* * *

Thus we now have the arc of the arbitrary angle divided into six (2×3) parts.

* * *

Since we are only interested in the division of the arbitrary angle in excess of 90° into three parts, we then take every other division point or the first one adjacent to the central bisector of the arbitrary angle. Doubtlessly it is now clear how and on what principle it is essential to divide any arbitrary angle up to 360°. That would be the second and final part of the trisection scheme “B” for an arbitrary angle with a compass and unmarked straightedge ruler.

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