Euclid's Elements Book XIII Proposition 18
In a polygonal pyramid, the angle between a base edge and a lateral edge (edge connecting the base to the apex) is called the base-edge angle or edge angle. It's one of several angles you can consider in a pyramid, including angles between faces (dihedral angles).
Relating the polygonal pyramid edge angle to a roof-framing hip rafter is the easiest way to prove Euclid's Book XIII Proposition 18 is correct. As well, as my drawing of Proposition 18. Euclid states that the line HC is 5 times the square of lines HK and KC. So, how did Euclid know this was true? How do we prove it's true. How do we know the lines representing the edge length of the Platonic Solids are the edge lengths of the Platonic Solids when the Platonic Solid is circumscribed in a sphere with a radius of one?
For some reason, Euclid didn't use Euclid's Elements Book II Proposition 11 Extreme and Mean Ratio for The Golden Ratio as the base drawing for Proposition 18. As, seen on the next couple of pages the Extreme and Mean drawing provides all the proof he needed for comparing the edge lengths of the Platonic Solids when inscribed in a sphere with a radius of one.
Radius EK = 1
AN side of Cube = 1.15470
AK side of Octahedron=1.41421356
AM side of Icosahedron = 1.051462
CN side of Tetrahedron = 1.632993
AQ side of Dodecahedron = 0.713644
Icosahedron Development
Line AM is the edge length of the Icosahedron inscribed in a sphere with a radius of 1. Point E is the center of the sphere. Points A-M are points on the sphere. Line SM is a plane perpendicular to the line AE. Line SM is the dimension to the center of the deck-plan view drawing of the pentagon with a sphere that has a radius of 1.
Line AH is the extreme ratio of the Golden Ratio. Angle ACH is the Golden Slope Triangle. Line EB is the square root of 5. Both lines CH and EB intersect at point M. Therefore, we can say the Edge of the Icosahedron is developed from the Golden Ratio and the square root of 5.
Line AQ is the edge length of the Dodecahedron inscribed in a sphere with a radius of 1. Point E is the center of the sphere. Points A-Q are points on the sphere. Line YQ is a plane perpendicular to the line AE. Line YQ is the dimension to the center of the deck-plan view drawing of the Equilateral Triangle with a sphere that has a radius of 1.
Line AH is the extreme ratio of the Golden Ratio. Angle GCF is developed from the extreme ratio of the Golden Ratio. Line XB is developed from the Golden Ratio of 1.61803. Both lines GC and XB intersect at point Q. Therefore, we can say the Edge of the Dodecahedron is developed from the Golden Ratio and the extreme ratio.
To prove the lines forming the triangles containing the edge length in the semi-circle of the sphere are correct, you can develop the deck-plan view for each of the Platonic Solids as a roof framing exercise. Then, develop the Edge angle triangle(hip rafter slope triangle) and use a bevel square to compare the angles in the triangle to the angles in the semi-circle.
The deck plan view drawing is developed from a vertex point and the face that intersects the vertex point.
Tetrahedron
is defined by four equal equilateral triangles and four Vertices
with three edges meeting at each vertex
Cube
is defined by six equal squares and eight Vertices
with three edges meeting at each vertex
Octahedron
is defined by eight equal equilateral triangles and six Vertices
with four edges meeting at each vertex
Icosahedron
is defined by twenty equal equilateral triangles and twelve Vertices
with five edges meeting at each vertex
Dodecahedron
is defined by twelve equal equilateral triangles and twenty Vertices
with three edges meeting at each vertex
To complete the proof of my drawing for proposition 18, you can draw out the hip rafter slope triangles using any radius for the semi-circle. The edge length of the Platonic Solid in the semi-circle is equal to 2 times the radius of the semi-circle times the "rise" leg of the hip rafter slope triangle when the hypotenuse of the hip rafter slope triangle is equal to the radius of the sphere. This is supported by using congruent angles for the edge length of the Platonic Solid.
A couple of extra drawings of the Icosahedron with the √5 triangles that develop point M.
2 x Radius of the sphere x gd = Edge Length in the sphere
Icosahedron developed from √5 Triangle.
Icosahedron Developed from Five Squared
Euclid states that HK is quadruple of the square at KC. Then he says that the square on HK is five times the square of HK and KC. This is the same for line aE that develops the Golden Ratio of 1.618.
As, seen in this drawing the line AF ,Golden Ratio, is also the straight line of a Hexagon & Decagon. Then we can use the line AF, Golden Ratio, to develop the deck-plan view of the pentagon of the Icosahedron. Line HK is the radius of the deck-plan view of the pentagon and line AH is the length of each side in the pentagon, as well as each side of an Icosahedron.
Only an Alien would have known that cutting the line AL into extreme and mean ratio would have developed the Golden Ratio of AX that is part of the straight line containing the Hexagon and Decagon inscribed in a circle with the radius equal to one.
The vertical lines EC,FD, ML, and QH are the radius's used for the development of the deck-plan view drawings. These radius's develop the Platonic Solid with a sphere circumradius of 1.
