# Golden Section in Two Equilateral Triangles, II

## Quang Tuan Bui

### November 7, 2011

ABC and AMN are two equilateral triangles where M is midpoint of side BC. Arc 60° centered at B passing through A, C intersects side MN by golden ratio.

Construct two circles: circle A(P) centered at A passing through P and circle C(A) centered at C passing through A. These two circles intersect at two points. Let F denote the one that is in the same side of AC as N.

Suppose D is symmetric to A in M. Two triangles ABP and ACF are congruent because

The rotation R(A, 60°) also maps ΔAMP onto ΔANF; therefore,

(1) | MP = NF |

and ∠FNA = ∠PMA = ∠MAN = 60°, implying

(2) | FN||AD. |

∠APD = 120° and ∠APF = 60°, so that

(3) | D, P, F are collinear. |

From (2) and (3) two triangles PMD, PNF are similar. Therefore:

NF/PN = MD/PM

MP/PN = NF/PN by (1)

Now we can calculate:

x | = MP/PN = MD/PM = AM/PM = NM/PM | |

= (NP + PM)/PM = NP/PM + 1 = PN/MP + 1 |

or x = 1/x + 1. From this x = φ, the Golden ratio.

(There is another proof for the same construction.)

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