Inverse proposition: A triangle is an isosceles triangle if the median lines of its two waists are equal.

If established, it is a true proposition.

Prove: In the triangle ABC, let the AC midline BD and AB midline CE.

Even DE, extend BC to f, make CF=DF, even DF.

Because d and e are that midpoint of AC and AB, respectively.

So DE is the median line in triangle ABC.

So de//cf;

Because DE=CF

So DECF is a parallelogram.

So CE//DF

CE=DF

Because BD=CE

So BD=DF

So ∠F=∠DBC

Because CE//DF

So ∠F=∠ECB

So ∠ ECB =∠DBC.

Because CE=BD

∠ ECB =∠DBC

BC=CB

So the triangle DBC is identical to the triangle ECB.

So BE=DC

Because BD and CF are the center lines.

So AB=2BE.

AC=2DC

So AB=AC

Triangle ABC is an isosceles triangle.