Golden Ratio and Midpoints

In a 2011 article, Jo Niemeyer, offered an elegant way of constructing the Golden Ratio with three equal segments, their midpoints and a pair of perpendicular lines.

Three equal segments A1B1, A2B2, A3B3 are positioned in such a way that the endpoints B2, B3 are the midpoints of A1B1, A2B2 respectively, while the endpoints A1, A2, A3 are on a line perpendicular to A1B1.

In this arrangement, A2 divides A1A3 in the golden ratio, namely, A1A3 / A1A2 = φ.

For a proof, drop a perpendicular B3H from B3 to A1A2.

Assume for convenience that all three line segments are of length 2. Then in right triangle A3B3H, A3B3 = 2, and B3H = 1/2 (as a midline in ΔA1A2B2).

By the Pythagorean theorem,

(A3H)² = (A3B3)² - (B3H)² = 4 - 1/4 = 15/4,

so that A3H = 15/2.

On the other hand, in right triangle A1A2B2, A2B2 = 2, and A1B2 = 1, making A1A2 = 3 and A1H = 3/2.

It follows that A1A3 = A1H + A3H = 3(5 + 1)/2, and, finally, A1A3 / A1A2 = [3(5 + 1)/2] / 3 = φ.


  1. Jo Niemeyer, A Simple Construction of the Golden Section, Forum Geometricorum, Volume 11 (2011) 53
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