Prove that $\sqrt{2}$ is irrational using only geometric concepts and proofs. The proof should look like a proof in Euclid's elements or standard high school geometry. No algebra is allowed.
(I know one proof - I am interested in seeing how many there are.)
On
Not exactly what you're looking for, but I think it is worth mentioning.
The argument used in this article by Marty Ross and Burkhard Polster gives a geometric interpretation to the standard proof by contradiction. They attribute the idea to John Conway.
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Here it is in Euclid's Elements (I assume you want it in translation, not in the original Greek):
http://aleph0.clarku.edu/~djoyce/java/elements/bookVIII/propVIII8.html
Let $\triangle ABC$ be an isosceles right triangle, with hypotenuse $|BC|=a$ and legs $|AB|=|AC|=b$. If $\sqrt{2}$ is rational, we can properly scale our triangle to assume that $a$ and $b$ are integers with $\gcd(a,b)=1$.
Consider the point $D$ on the hypotenuse that is a distance $b$ away from $B$, and draw a perpendicular from this point to $\overline{AC}.$ Let $E$ be the point where the perpendicular meets $\overline{AC}$. Then by similarity of triangles $\triangle ABC\sim\triangle DEC$, we have $|DC|=|DE|=a-b$ and by congruent triangles $\triangle ABE\cong\triangle DBE$ we have $|AE|=a-b$ so that $|EC|=2b-a$. It follows again by similar triangles that:
$$\frac{a}{b}=\frac{2b-a}{a-b}$$
This ratio contradicts $\gcd(a,b)=1$.