A geometric proof of the irrationality of $\sqrt{2}$ works by constructing two right isosceles triangles with legs $n$ and hypotenuse $m$, and finding in the construction similar triangles with legs $m-n$ and hypotenuse $2n-m$. This shows that $m$ and $n$ can't both be integers as it would lead to infinite descent.
I realise this is really a proof that the hypotenuse-leg ratio of a right isosceles triangle is irrational, and doesn't use the fact that this ratio is equal to $\sqrt{2}$, but that's an aside.
I've been trying to find a geometric construction to prove the irrationality of $\sqrt{3}$ in a similar way. I would expect this to involve 90°–60°–30° triangles. But I keep hitting dead ends.
Can it be done?
With a figure like the one on the left, Apostol demonstrated that an $(n,n,m)$ right triangle gives rise to a necessarily-smaller $(m-n,m-n,2n-m)$ triangle. Thus, there is no smallest triangle with integer sides, hence no such triangle at all, implying that $m/n=\sqrt{2}$ cannot be rational.
Likewise, the figure on the right shows that an $(n,m,2n)$ right triangle yields a necessarily-smaller $(2n-m,2m-3n,2(2n-m))$ triangle, so that $m/n=\sqrt{3}$ cannot be rational.