# 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

### References

- M. Bataille,
__Another Simple Construction of the Golden Section__,*Forum Geometricorum*,Volume 11 (2011) 55

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