From Equilateral Triangle and Square to Golden Ratio
In a 2011 article, M. Bataille, offered an elegant way of constructing the Golden Ratio via an equlateral triangle and a square.
Given an equilateral triangle ABC, erect a square BCDE externally on the side BC. Construct the circle, center C, passing through E, to intersect the line AB at F. Then, B divides AF in the golden ratio.
For a proof, drop a perpendicular CH from C to AB.
Assume for convenience that AB = 2, then CH = √3 and
FH² = CF² - CH² = 8 - 3,
so that FH = √5, making
- M. Bataille, Another Simple Construction of the Golden Section, Forum Geometricorum,Volume 11 (2011) 55