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.
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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 $.
Is this correct so far? How do we continue?
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Here is a diagramm of a book:
Under the assumption that the sides are commensurable, you can multiply all the lengths by an integer factor and assume that $a,b$ are integers. Then the construction gave you a new triangle $A_1C_1D$ (I am looking at the diagram on the left) that is similar to the original one, and that also has integer sides $2a-3b$, $2b-a$, and $2(2b-a)$.
Since $b<a$ and $2b>a$, it follows that $0<DA_1=2b-a<b=AD$. The construction, if you were to repeated with this new smaller triangle, can be repeated indefinitely, strictly reducing the side opposite to the angle of $30$ degrees, and keeping all sides positive integers. That is impossible.
A similar construction, that it looks simpler to me:
As in your picture call $b=AD$, $a=CD$, and $CA=2b$. Take $D_1$ on $AC$, with $CD_1=CD$. Then $D_1A=2b-a$. Lift a perpendicular to $AC$ at the point $D_1$ that intersects $AD$ at $F_1$. Since the triangles $ACD$ and $AD_1F_1$ are similar, then $F_1A$ is the double of $D_1A$. Therefore, $F_1A=2(2b-a)$. Subtracting we get that $DF_1=b-2(2b-a)=2a-3b$. Since $F_1D$ and $F_1D_1$ are tangents to the circle with center $C$ and radius $CD$, and both start from the same point $F_1$, then $F_1D_1=F_1D=2a-3b$.
We have obtained a new triangle $AD_1F_1$ that again has integer sides. The side opposite to the angle of $30$ degrees is now $2b-a$ which is positive and strictly less than the one in the original triangle $ACD$, which was $b$.