Conjecture about two triangles with equal areas

Question

About ten days ago, I made a conjecture about equal-area triangles while playing with GeoGebra, but I couldn't prove it.

If we have two triangles of equal area, then the ratio between the sum of the dimensions of the vertices of the first triangle from the midpoint of the centers of the two triangles and the sum of the dimensions of the vertices of the second triangle from the midpoint of the centers of the two triangles is limited to the range $[\frac{1}{2},2]$.

In other words: Let $∆ABC$ be a triangle and $M$ be the point of intersection of its medians, and let $∆abc$ be a triangle and $m$ be the point of intersection of its medians, and its area equal to the area of ​​triangle $ABC$ and let $O$ be the midpoint of the two points $M$ and $m$, then the property states the following:

$\frac{1}{2}≤\frac{AO+BO+CO}{aO+bO+cO}≤2$

In fact, what needs to be proven is only one of the two terms, and the other is produced directly by exchanging roles between the two triangles

If anyone can prove this, please do so

Also, if there are references indicating that this feature was previously discovered, please post them in the comments.

Answer 1

We give you a counter example, for which the ratio in question exceeds 2.

Answer 2

You can stretch one of the trangles by shifting two of its vertices in opposite directions, in such a way that its area stays constant, and $m$ and $M$ remain close to each other. Then the respective sum becomes arbitrarily large and the ratio tends to zero or to infinity.

FIX: ...its area stays constant, as well as its centroid (the intersection of its medians).