Math fact family number triangles

$A+C+E=68$, and the numbers at the corresponding vertices of the second figure also sum to $68$. Henceforth I’ll use $A'$ for the vertex in the second figure that corresponds to vertex $A$ in the first, so $A'=24$. $B+E=34$, which is exactly half of $68$, and it seems unlikely to be a coincidence that $C'+F'$ is also $34$. We may tentatively conjecture that $C+F$ and $A+D$ are also supposed to be $34$, so that $F=11$ and $D=9$. Similarly, $B+G=F'+H'=17$, exactly half of $34$, so we may guess that $I=17-D=8$ and $H=17-F=6$. The second figure can be completed similarly. Note that $B+D+F=34$ and $G+H+I=17$. Thus, we have a clear pattern.

And on further examination, we find that it is to a considerable extent composed of triangular ‘math fact families’:

• $9+11=20$ for the top triangle;
• $9+14=23$ for the lower right triangle;
• $11+14=25$ for the lower left triangle;
• $3+8=11$ for the upper left blue triangle;
• $3+6=9$ for the upper right blue triangle; and
• $6+8=14$ for the bottom blue triangle.

This is not entirely satisfactory, since the yellow triangle in the middle is not of this type, and I’d also be happier if the number at the bottom of the problem were $136=2cdot 68$, but I can’t see anything better at the moment.