prove that $\sqrt{2}$ is irrational using only geometry 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.)
 A: 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 $\gcd(a,b)=1$.
A: 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.
A: 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
