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These triangles each have all three of their sides of equal
length, and such triangles are referred to as being 'equilateral'.
Another property of equilateral triangles is that all three of their
angles are also equivalent. Furthermore, each of these equal angles
contains 60 To facilitate further discussion, an enlargement of one of the above equilateral triangles is presented in Diagram 7. A new line (AM) has been drawn from the center of the circle to a side of the hexagon, to meet it at a right angle. As can be satisfactorily proven through empirical measurement, the addition of this line divides the equilateral triangle directly in half, creating two equal right triangles. Since this means that the line MB equals line MC, each must be half of a unit in length (BC being originally equal in length to one radius). This then means that we now know the length of two of the three sides of each of these two right triangles (AB and AC = 1, and MC and MB = 1/2).
The Pythagorean Theorem then allows the derivation of the length of the third side (AM in the diagram) to be easily achieved.13 AM2 + (1/2)2 = (1)2 This leads to: AM2 = 1 - 1/4 = 3/4 And so, AM = An ancient Egyptian researcher could now have understood that a
right triangle whose height and base were in the ratio of Furthermore,
from studying the manner in which a square's diagonal divides a
square in half (see Diagram 2 in previous section), he would note
that in that instance the ratio of the height to the base would be 1
to 1, and that the base angles accompanying such a ratio would be 45 Diagram 8 shows that with a little extrapolation there is still more that can be straightforwardly gleaned from the inscribed hexagon.
Line
AM has been extended to touch the circle (at D). Lines BD and CD are
drawn in, and a new perpendicular is dropped from point A to meet BD
at point E. As before, we have a situation where two equivalent
right triangles are formed, Is there now a way to find the comparative lengths of the line segments AE and ED? Indeed there is. Triangle BMD is a right triangle, and we already know that MB equals 1/2 the length of the radius. Also, it is clear that MD equals AD minus AM. Therefore: MD = 1 - The Pythagorean Theorem now can be used to solve for BD: BD2 = MB2 + MD2 = (1/2)2 + (.133975)2 = .267949, and so BD = .517638 ED is then half of this, and with this knowledge, the length of AE can be found, again by using the Pythagorean Theorem. (It turns out to be .96593. Bear in mind that all of these results would have been computed, and expressed, utilizing the Egyptian unit fraction method. This is a task within ancient Egyptian computational capability). It
is at this point that the Egyptian priests may well have felt that
they were very much on to something, considering that they would
have now computed right triangle ratios for angles of 15 Certainly more work would have been done to see what else could be extracted from diagrams which stem from the inscribed hexagon.
If the scribe had set his dividers to subtend the length of the equilateral triangle's perpendicular (line AM in Diagrams 7 and 8, and blue line in Diagram 9), he would have found that it appears to mark off a perfect 7 equal-sided figure (heptagon) around the inside of the circle. AM is, in fact, .001715 smaller than the side of a truly accurate equal-sided inscribed heptagon, but this deficiency is not readily detectable empirically. The inscribed heptagon would have allowed for the nearly exact computation of the sides of right angle triangles associated with 1/14th, 1/28th, and 1/56th, etc., of a full rotation. (Because AM, whose relative length is .8660254, is only a close approximation to the side of a true inscribed heptagon, computations based on this length would necessarily be off by a correspondingly small amount). Furthermore, as explained in the Appendix, the inscribed hexagon can also provide a means (though a slightly more complex one) for deriving a very nearly accurate inscribed eleven equal-sided figure (an "endecagon"), with associated computable right triangles.14 It is my opinion that the heptagon and endecagon would not have played an essential role in the either the derivation of trigonometry nor in the design process of the Great Pyramid.15 I say this largely because there was a much more elegant means at hand for expanding trigonometric capabilities, and also because this latter method arises from a numerical constant whose existence in nature may possibly have been seen as being even more mysterious than that of the Pi phenomenon. Next Section: Pentagon and Trigonometry 13.
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