A Robbins Pentagon bound to any (non-isosceles) Integer Triangle?

Question

Given any non-isosceles triangle $\triangle ABC$, and denoting $AB$ its longest side, the following construction

determines the points $DFGE$ (see this post for details).

My conjecture is that if $\triangle ABC$ is an integer triangle, then the pentagon $DFCGE$ is a Robbins pentagon, and viceversa.

I am not sure of this claim. Therefore, I ask your advice both to assess if this is true and, otherwise,

to find the supplementary conditions needed.

Thanks for your help!

Answer 1

No, let $AD=x=2$ and $EB=z=1$ and $DE = y=2$. Then

$$\begin{eqnarray}c=x+y+z &=& 5\\ b=x+y &=& 4\\ a=y+z &=& 3\end{eqnarray}$$

Now, $$ \cos \alpha = {b^2+c^2-a^2\over 2bc} = {4\over 5}\in\mathbb{Q}$$ so $$DF^2 = 2x^2-2x^2\cos \alpha = x^2{2\over 5}\implies DF\notin \mathbb{Q}$$