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Solution to puzzle 91: Nonagon diagonals

In regular nonagon ABCDEFGHI, show that AF = AB + AC.

Regular nonagon ABCDEFGHI, showing diagonals AC and AF.
Regular nonagon ABCDEFGHI, showing diagonals DH and EG. DE and HG are continued, to meet at X.

Continue DE and HG to meet at X.

By symmetry, AC = EG and AF = DH.
Also by symmetry, EG and DH are parallel to AB.

Line segment BC makes an angle of 360° / 9 = 40° with line segment AB (or its continuation.)
Hence CD makes an angle of 80° with AB, and DE makes an angle of 120° with AB.

Hence angleHDX = 60°.  Similarly angleXHD = 60°, and so angleDXH = 60°.
It follows that triangleXHD and triangleXGE are both equilateral.

Hence DH = DX and EG = EX.
So, DX = DE + EX implies DH = DE + EG.

Therefore, in regular nonagon ABCDEFGHI, AF = AB + AC.

Alternative proof

Madhukar Daftary sent the following elegant solution.

Draw diagonals ID and IG.  Let AF and ID intersect at O.

Regular nonagon ABCDEFGHI, showing diagonals AC, AF, ID, and IG.

By symmetry, CD, AF, and IG are parallel.  Similarly, AC and ID are parallel.

Also by symmetry,

AO = CD = AB
OF = IG = AC

Adding, we obtain

AF = AO + OF = AB + AC

Heptagon diagonals (3 star)

In regular heptagon ABCDEFG, show that 1/AB = 1/AC + 1/AE.

Regular heptagon ABCDEFG, showing diagonals AC and AE.
Hint  -  Solution

Further reading

  1. Nonagon
  2. Regelmäßiges Neuneck

Source: Original

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