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- Apr 14, 2013

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I want to prove the incommensurability at an equilateral triangle.

The way of such a proof is the following:

1) We are looking for a "common measure" of two lines $ a $, $ b $, i.e. a line $ e $ that measures both $ a $ and $ b $ integer (i.e. there are natural numbers $ m $, $ n $with $ a = me, b = ne $).

2) We take the shorter of the two parts (e.g., $ b $) away from the longer (e.g., $ a $) until the remaining piece $ r_1 $ is shorter than $ b $. We're taking away $ r_1 $ from $ b $ as often as ... and so on.

3) If the process breaks down, there is a common measure, the stretches are then commensurate; otherwise incommensurable.

I found the proof (in german) here (pages 941-942) but I haven't really understood the method.

What I understand, is the following:

We consider an equilateral triangle ABC with side length $a$. Let CD be the height of the triangle from $C$.

We assume that $ AD $ and $ CD $ have a common $ e $ measure.

Then, for appropriate $ m $ and $ n $: $ AD = ne $ and $ CD = me $.

Now we take the shorter of the two stretches, $ AD $ off the longer, $ CD $, until the remaining $ r_1 $ is shorter than $ AD $.

To do this, we take a circle around the point $ D $ with radius $ AD $, so this intersects the height $ CD $ at point $ E $.

We have that $CE=CD-ED=CD-AD=a-b$

We draw then a circle with center $E$ and radius $CE=a-b$. This intersects the segment $CA$ at the point $F$.

We have that $EF=CE=a-b$.

We draw a parallel line to $BC$ from $F$. This intersects the segment $AD$ at $A_1$.

The triangle $AFA_1$ is equilateral, since $FA_1$ is parallel to $BC$ and so all the angles of $AFA_1$ are equal to $60^{\circ}$.

We construct an equilateral triangle $A_1B_1C_1$ with side length $2\cdot A_1D$.

Is everything correct so far? Or have I understood the construction wrong?

What I don't understand how we can determine the length of the sides of the new triangle. Could you explain to me how we can find these values?