Making a nonagon with ruler and circle 1
Take a point as midpoint for a circle with radius r = a.
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1
2
Take on the circumference a point as being the midpoint for another circle with a radius a. r = a. The two circles intersect at points A and B.
A
B
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3 Take intersection point A as midpoint for a circle with radius a. The circle intersects the other circles at points C, D, E and F.
A
D
E
C
F
B
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3
4 Take intersection point B as midpoint for a circle with radius a. The circle intersects the other circles in the already existing intersecting points E and F and newly in the points G and H.
A
D
E
H
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C
F
B
G
4
5 Take intersection point G as midpoint for a circle with radius a. The circle intersects the 4 present circles in the intersecting points B and F and newly in the points i and J.
A
D
C
i E
F
H
G B
J
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5
6 Take intersection point H as midpoint for a circle with radius a. The circle intersects the 4 present circles in the intersecting points B and E and newly in the points K and L.
A
D
C
L
i E
F
H
G B
K
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J
6
7
Draw a line from point A to B.
A
D
C
L
i E
F
H
G B
K
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J
7
8
Draw a line from point G to E.
A
D
C
L
i E
F
H
G
B
K
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J
8
9
Draw a line from point H to F.
A
D
C
L
i E
F
H
G B
K
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J
9
10
take the intersecting point of the 3 lines as midpoint N of a circle with a radius equal to line ND.
A
D
C
L
i N
E
F
H
G B
K
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J
10
11 Extend the lines NA to NA2, NG to NG2 and NH to NH2 and draw the lines NC , Ni , NJ , NK and NL. These are the radii of the nonagon.
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
11
12
The length of the edges of triangle BEF is equal to the edges of the nonagon.
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
12
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
© Jos Hoebe.
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J
13
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13 The length of the edges of triangle AGH is equal to the length of the edges in the order of their appearance: A2i, iJ, JH2, H2D, DC, CG2, G2K, KL and LA2 by which a nonagonal star appears als two interwoven pentagons.
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
15
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
16
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14 The length of the edges of triangle A2G2H2 is equal to the edges of the triangles CJL and iKD, by which a nonagonal star appears of three the same midpoint encircling triangles.
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
18
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
19
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15 Trough the combination of the length of the edges of the triangles AGH and A2GcH2 appears a often used nonagonal figure known as the enneagram.
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
21
A2
D
C
A
i
L E
F N
H
B
G
H2
G2
K
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J
22
9
8
1
2
7
6
3
5
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4
23
16
Take length AB
D
C
A
i
L E
F
H
B
K
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G
J
24
17
take N as midpoint for a circle with a radius r = AB so that radius NA3 originates.
A3
D
C
A
i
L E
F N
H
B
K
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G
J
25
18
And NG3 and NH3
A3
D
C
A
i
L E
F N
H
B
G
H3
G3
K
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J
26
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20 The length of the edges of triangle A3G3H3 is equal to the length of the lines A2J – JD – DG2 – G2L – Li – iH2 – H2C – CK – KA2 in this order connected to each other in circle A2 – C – i - G2 – J – K - H3 – L – D - A2 so that a nonagonal star appears.
A3
D
C
A
i
L E
F N
H
B
G
H3
G3
K
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28
A2
D
C
A
L
i E
F N
H
B
G
H2
G2
K
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J
29
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30
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31
The nonagon and its circumscribed circle and the three possible nonagonal stars features as combination the following figure:
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32
The 4 used triangles, the 3 used circles and the 3 centrifugal radial lines as a combination is this figure:
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