Are there Heronian triangles that can be decomposed into three smaller ones?

Yes, for example the 13-14-15 triangle can be scaled by 11 to find a point $D$ at distance $80$, $91$, $102$ from the vertex opposite the side of length $11 \cdot 13$, $11 \cdot 14$, $11 \cdot 15$ respectively:

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This was found by fixing $A,B,C$ and searching through the points $D$ of low height, using the parametrization of Pythagorean triples $x^2+y^2=d_1^2$ to search through only those $D$ that are already at rational distance from $(0,0)$, and then testing whether the distances to $B$ and $C$ are rational as well.

In fact, for any choice of triangle $ABC$, such points $D$ should be dense in the Euclidean plane, and thus in the interior of the triangle, because they're parametrized by a K3 surface with enough structure that standard tricks apply.

Let $A,B,C$ be the vertices of any rational Heronian triangle. We may choose Euclidean coordinates so that $A,B,C = (x_i,y_i)$ ($i=1,2,3$) with all $x_i$ and $y_i$ rational (for example, put $A$ at the origin and $B$ at $(x_2,0)$, etc.). Then the points $D=(x,y)$ at rational distances $d_i$ from $x_i$ correspond to solutions $(x,y,d_1,d_2,d_3)$ of the three Diophantine equations $(x-x_i)^2 + (y-y_i)^2 = d_i^2$ with each $d_i$ positive. Thus we seek rational points on the intersection $S$ of three quadrics in 5-space. Here the singularities of this intersection are mild enough that $S$ is birationally a K3 surface, as it would be if the intersection were smooth.

The geometry yields several elliptic fibrations on $S$; e.g. for any line through (say) $A$ whose slope comes from a Pythagorean triangle, the points $D$ on that line that that are also at rational distance from $B$ and $C$ are parametrized by a genus-$1$ curve with rational points at infinity. Starting from those rational points (or those for which $D$ is the orthocenter, or indeed one of the vertices $A,B,C$), it should be straightforward to bounce around a few elliptic fibrations to find a dense set of rational points.

There should be many examples where the interior vertex lies on the perpendicular bisector of one of the sides. It's best to redefine a Heronian triangle to be one with rational sides and rational area. I'll call such a T "standard" if its vertices are at $(-2,0), (2,0)$ and $(r,s)$ where $r$ and $s$ are rational. Every Heronian triangle is similar to a standard one.

Now let T be standard and P be $(0,(x)-(1/x))$ where $x$ is rational. Then P has rational distance from 2 vertices of T. The condition that it have rational distance from the third is that there exists a rational y such that $(xx-sx-1)^2 +(rx)^2 =y^2$. If the elliptic curve one gets in this way has positive rank then there will be a dense set of points $(0,(x)-(1/x))$, all lying on the perpendicular bisector of the base of T, each giving the desired decomposition of T.

I worked out the case $r=-2, s=3$. Unfortunately the curve one gets is one of conductor 15 with 8 rational torsion points and rank 0. But there must be lots of choices of r and s where the rank is positive.

EDIT: Here's another construction which should give many examples where the interior point lies on an altitude. Consider a Heronian triangle with the base extending from $(0,0)$ to $(a+b.0)$, and the foot of the altitude to the base at $(a,0)$. Let P be $(a,x)$ where $x$ is rational. Then P is at a rational distance from one vertex, and is at a rational distance from the other two when there are rational u and v with xx+aa=uu and xx+bb=vv. These equations again define an elliptic curve and one will get a dense set of points (a,x) on the altitude, each giving a desired decomposition, when the curve has positive rank.

The interesting question then seems to be the existence of a point, that lies neither on an altitude nor on the perpendicular bisector of a side, and that yields the desired decomposition.