Conjecture on equivalence of irrational-angle tilings to triquadratic tilings

Let $ABC$ be a triangle that is $N$-tiled, meaning it is tiled by $N$ congruent triangles. Let the tile have angles $\alpha$ and $\beta$ satisfying

$$3\alpha+2\beta=\pi,$$

and suppose that $\alpha$ is not a rational multiple of $\pi$. Equivalence conjecture. The tiling is equivalent to a triquadratic tiling, where equivalence is the transitive closure of replacing a parallelogram in a connected component by a parallelogram lattice-tiled in a different way. This concerns the unresolved classification of tilings and the number of tilings for a fixed triangle, tile, and $N$.