Five Similar Triangles from 4-5-6

A 4-5-6 triangle can be divided into 5 similar scalene triangles. I have the solution, but I figured out it was a 4-5-6 after putting together the component triangles. Can anyone else find it?

The Zak triangle can be divided into 6 similar triangles in 2 ways. It leads to an interesting fractal.

I have a collection of unusual similar triangle polygons. I believe only three quadrilaterals can be divided into 4 similar obtuse triangles with a common vertex. One is used in the 4-5-6, the Zak triangle uses $\rho$, the plastic constant, the third uses $\psi$, the supergolden ratio. Below the numbers are powers of $\sqrt\psi$. This quadrilateral leads to interesting fractals.

What other triangles can be divided into similar obtuse triangles in non-trivial ways?

This is related to New Substitution Tilings Using 2, φ, ψ, χ, ρ.

For the secondary question of the 4-5-6 triangle:

Start with $\triangle ABC$ with $|AB|=4, |AC|=5, |BC|=6$. Bisect $\angle A$ and $\angle B$, drawing their rays to their point if intersection at the in center $I$. From $I$ construct ray $ID$ parallel to $AC$ where $D$ lies on $BC$. Construct point $E$ on $AC$ such that $|AE|=4$ or equivalently, $IE$ is congruent to $IB$. Alternatively, place $E$ so that $|AE|=1$ or $DE$ is congruent to $IB$ (this alternative is shown in the diagram below). Draw segments $IE$ and $DE$ to complete the dissection.

I have not formally proved it, but it seems to work.

The existence of two solutions for the 4-5-6 triangle, with either $|AE|=1$ or $|AE|=4$, implies a fourth quadrilateral divisible into four similar obtuse triangles with a common vertex. Two of the quadrilaterals are the possible $ABDE$ quadrilaterals from the 4-5-6 triangle solutions, the other two are the quadrilaterals based on the cubic roots as given by the OP.