Why doesn't an isosceles triangle belong to the projective plane?

Question

I am reading Coxeter's projective geometry book and he says that the isosceles triangle doesn't belong to the projective plane. I see how in projective geometry we solely use the compass as opposed to in Euclidean geometry we use both the compass and straightedge. In projective geometry do we only use the compass for construction and in Euclidean geometry we use just the compass and straightedge for construction but is it the same as in Euclidean geometry where not everything in the plane is constructible? Because since the projective plane is really the Euclidean plane + line at infinity there is nothing stopping from an isosceles triangle existing even though we can't measure anything with a ruler.

Answer 1

If you fix an embedding of $\mathbb{A}^2(\mathbb{R})$ (affine plane) in $\mathbb{P}^2(\mathbb{R})$ (projective plane), and fix some euclidian structure on $\mathbb{A}^2(\mathbb{R})$, then of course you can take an isoceles triangle in $\mathbb{A}^2(\mathbb{R})$ and consider its image in $\mathbb{P}^2(\mathbb{R})$. It exists.

But it just doesn't make any sense to call it isoceles anymore because there is no notion of length in $\mathbb{P}^2(\mathbb{R})$. For instance, this property won't be preserved by the projective automorphisms.