Golden Ratio Triangles and the Regular Pentagon

Click the buttons to show the steps in the proof.

Start Let triangle ABC have AB = AC = τ, BC = 1. Then ∠A = 36°, ∠B = ∠C = 72°. Here's how we know this:

Step 1 Let D be on AC, with AD = 1, so that DC = τ - 1 = τ-1.

Step 2 Then triangle BCD is similar to triangle ACB, because they have a common angle at C, and the adjacent sides are in proportion, that is,

BC/CD = AC/CB.
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Step 3 Therefore BD/BC = AB/AC; that is, BD = BC = 1.

Step 4 It also follows from Step 2 that ∠A = ∠DBC.

Step 5 From Step 3, triangle ABD is isosceles, so ∠A = ∠ABD.

Step 6 From Steps 4 and 5, ∠ABC = 2∠A.

Step 7 Because triangle ABC is isosceles, ∠ACB = 2∠A.

Step 8 The angles of triangle ABC add up to 180°, so 5∠A = 180°, so ∠A = 36°; and ∠B = ∠C = 72°.

A Corollary Triangle ABD, with sides τ, 1, and 1, has angles 36°, 36°, and 108°.

Regular Pentagon On AB and AC, construct triangles ABE and ACF congruent to triangle ABD, as shown in the figure. Then AFCBE is a regular pentagon.

Constructing τ | Some trigonometrical ratios

Last modified on $Date: 2015-04-23 17:26:33 -0400 (Thu, 23 Apr 2015) $

Christopher J. Henrich