Hexagonal Construction

825.21 Diameter: The Greeks then started another independent investigation with their three tools on the seemingly flat planar surface of the Earth. Using their dividers to strike a circle and using their straightedge congruent to the center of the circle, they were able with their scriber to strike a seemingly straight line through the center of construction of the circle. As the line passed out of the circle in either direction from the center, it seemingly could go on to infinity, and therefore was of no further interest to them. But inside the circle, as the line crossed the circumference at two points on either side of its center, they had the construction information that the line equated the opening of the dividers in two opposite directions. They called this line the diameter: DIA + METER.

825.22 Now we will call the center of the constructed circle D and the two intersections of the line and the circumference A andB. That AD =DB is proven by construction. They know that any point on the circumference is equidistant from D. Using their dividers again and using point >1 as a pivot, they drew a circle around A; they drew a second circle using B as a pivot. Both of these circles pass through D. The circle around A intersects the circle around D at two points, C and C. The circle around B intersects the circle around D at two points, E and E. The circle around A and the circle around B are tangent to one another at the point D.

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825.23 They have now constructed four equilateral triangles in two pairs: ADC and ADC as the first pair, and DBS and DBE as the second pair. They know that the lines AC, CD, AC, and DC are all identical in length, being the fixed opening of the dividers and so produced and proven by construction. The same is true of the lines DE, EB, DE, and BE they are all the same. The Greeks found it a tantalizing matter that the two lines CE and CE, which lie between the vertexes of the two pairs of equilateral triangles, seemed to be equal, but there was no way for them to prove it by their construction.

825.24 At first it seemed they might be able to prove that the increments CE and CE are not only equal to one another, but are equal to the basic radius of the circle AD; therefore, the hexagon ACEBEC would be an equilateral hexagon; and hexagons would be inherently subdivisible into six 60degree equilateral triangles around the central point, and all the angles would be of 60 degrees.

825.25 There seemed to be one more chance for them to prove this to be true, which would have provided an equiangular, equiedged, triangularly stable structuring of areal mensuration. This last chance to prove it was by first showing by construction that the line ADB, which runs through the point of tangency of the circles A andB, is a straight line. This was constructed by the straightedge as the diameter of circle D. This diameter is divided by four equal half-radii, which are proven to be half-radii by their perpendicular intersection with lines both of whose two ends are equidistant from two points on either side of the intersecting lines. If it could be assumed that: (1) the lines CE and CE were parallel to the straight line ADB running through the point of tangency as well as perpendicular to both the lines CO and EE; and (2) if it could be proven that when one end between two parallels is perpendicular to one of the parallels, the other end is perpendicular to the other parallel; and (3) if it could be proven also that the perpendicular distances between any two parallels were always the same, they could then have proven CE = CD = DEDE, and their hexagon would be equilateral and equiradial with radii and chords equal.