Is there anything known about the existence of Heronian triangles ABC (i.e. with rational side lengths and rational area) that can be decomposed into three Heronian triangles ABD, BCD, CAD? Equivalently, is there a degenerate Heronian tetrahedron such that one vertex is in the interior of the triangle formed by the three other ones?

(Edit: wrong alternative formulation removed)

Note that it is easy to find solutions for the problem of dividing ABC into three triangles whose common vertex is not an inner point, but one of A, B, C. For example, take (36,30,30) and divide 36=11+14+11, so the two "chords" have length 25 and all triangles are Heronian.

Edit: sorry, I have posted this too quickly. In fact, all so-called *Bis triangles* in "On Triangles With Rational Altitudes, Angle Bisectors Or Medians (1999)" Buchholz' Thesis, i.e. Heronian triangles with 3 rational bisectors, satisfy the condition. Those triangles can be equivalently characterized by the condition that the sides and $AO,BO,CO$ are rational where $O$ is the incenter. (Then their area is necessarily rational, too). Equivalently, in the well-known Carmichael parametrization $a=n(m^{2}+k^{2}) ,b=m(n^{2}+k^{2}),c=(m+n)(mn-k^{2})$, both $m^{2}+k^{2}$ and $n^{2}+k^{2}$ must be squares.

Also, if $H$ denotes the orthocenter of a Heronian triangle, it is easy to see that $AH,BH,CH$ are always rational.

So the following question is more interesting:

Do there exist Heronian triangles ABC that can be decomposed into three Heronian triangles ABD, BCD, CAD where D is an inner point other than the incenter or the orthocenter of ABC?