Application of algebra to trisect an angle of 60 degree

The origins of trisection of an angle began around 500 BC. Many many great mathematicians tried their best to solve this problem but miserably failed. Trisection of an angle and doubling the cube were proved impossible by Pierre Wintzel in 1837, although their impossibility was already known to Gauss in 1800. Squaring the circle problem was proved to be impossible by Lindemann in 1882 [1-7]. Both Wintzel and Lindemann applied the laws of Galois fi eld theory and derived their results. It is to be noted that in quantum mechanics and super string theories the basics of abstract algebra particularly the laws of Lie groups are widely applied. It is a well known fact that the predictions of Einstein’s special and general relativity theories have been experimentally established. But there are published experimental tests which challenge these theories. i.e. in some experiments these predictions do not hold [8-18].


Introduction
The origins of trisection of an angle began around 500 BC.
Many many great mathematicians tried their best to solve this problem but miserably failed. Trisection of an angle and doubling the cube were proved impossible by Pierre Wintzel in 1837, although their impossibility was already known to Gauss in 1800. Squaring the circle problem was proved to be impossible by Lindemann in 1882 [1][2][3][4][5][6][7]. Both Wintzel and Lindemann applied the laws of Galois fi eld theory and derived their results. It is to be noted that in quantum mechanics and super string theories the basics of abstract algebra particularly the laws of Lie groups are widely applied. It is a well known fact that the predictions of Einstein's special and general relativity theories have been experimentally established. But there are published experimental tests which challenge these theories.
Similarly, the authors fi ndings also challenge abstract algebra. The authors never never and never question the consistent fi eld of abstract algebra. But the authors fi rmly believe that the laws of abstract algebra can not fetter the angle trisection.

Construction
Construct an equilateral triangle ABC in Figure 1. On the extensions of AB and AC, make BD=DE=AB and AC=CE=EF respectively. Join B and F & D and F. Bisect BD at G. Join G and F. With centre C radius CB describe an arc cutting BF at H. Join C and H and produce it till it meets EF at I and GF at J.
In triangles ABC and AEF all the sides are equal and all the angles are equal. In triangle BCH, since BC and HC are equal, the base angles are equal. By SAS correspondence triangles AGF and EGF are congruent. So, the angle at G is 90 degree. By SAS correspondence triangles BGF and DGF are congruent. So, angles GBF and GDF are equal, and angles BFG and DFG are equal. By SAS correspondence triangles ABF and EDF are congruent. So, angles AFB and EFD are equal, and angles ABF 014 https://www.peertechzpublications.com/journals/annals-of-mathematics-and-physics Citation: Sivasubramanian Let the sum of the interior angles of all Euclidean triangles is equal to 180 degrees. (2)

Discussion
For trial measuring angle e, 2f and i, we get that e=2f=i=20 degree. Describing an arc with center H and radius HC, it moves through F. So, e = 20 degree is consistent. In this work, we have not introduced or assumed any conjecture or hypothesis. Only we have applied one of the fundamental operations of number theory. ( i.e. addition ). So, beyond any doubt e = 2f = 20 degree is consistent.

Conclusion
The authors attempts will open the further attempts which may explore new and fascinating results.