Fix a side of the larger triangle as a base. The triangles that contribute to the bottom base with a side have sum of homothecy ratios is $1$, since the sum of their bases is equal to the length of the original triangle's base. We claim that the sum of the remaining homothecy ratios, those of the triangles with a side parallel to the base but not coinciding with the base, is $0$, from which the result follows. Indeed, consider each line segment parallel to the base but not inside the base determined by the partition (since the smaller triangles are homothetic to the original, it is clear that each triangle contributes to such a segment). On one side, we will have some (possibly just one) triangle, who's sums of homothecy ratios is $r$. On the other side of the segment, we must have triangles of opposite orientation that in total cover the same distance, and hence the same proportion of the base side length. The sum of the ratios for these triangles must be $-r$. Summing over all line segments, which is equivalent to summing over all smaller triangles, we get that the sum of homethecy ratios must be $1$.