Click the buttons to show the steps in the proof.
Let triangle ABC have AB = AC = τ, BC = 1.
Then ∠A = 36°, ∠B = ∠C = 72°.
Here's how we know this:
Let D be on AC, with
AD = 1, so that DC = τ - 1 = τ-1.
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,
Therefore BD/BC = AB/AC; that is,
BD = BC = 1.
It also follows from Step 2 that ∠A = ∠DBC.
From Step 3, triangle ABD is isosceles, so
∠A = ∠ABD.
From Steps 4 and 5, ∠ABC = 2∠A.
Because triangle ABC is isosceles,
∠ACB = 2∠A.
The angles of triangle ABC add up to
180°, so 5∠A = 180°, so
∠A = 36°; and ∠B
= ∠C = 72°.
Triangle ABD, with sides τ, 1,
and 1, has angles 36°,
36°, and 108°.
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