Equilateral triangles ABC and A1B1C1 are given. O and O1 are, respectively, the intersection points of the medians

Equilateral triangles ABC and A1B1C1 are given. O and O1 are, respectively, the intersection points of the medians of these triangles, OA = O1A1. Prove that triangles ABC = A1B1C1.

Since triangles ABC and A1B1C1 are equilateral, all their internal angles are 60, and their medians AM, A1M1, BH and B1H1 are also heights and bisectors.

Then the triangles AOH and A1O1H1 are rectangular, the angle OAN = O1A1H1. By condition, OA = O1A1, then triangles AOH and A1O1H1 are equal in hypotenuse and acute angle, and then AH = A1H1, and therefore AC = A1C1, since points H and H1 are the middle of AC and A1C1.

And since the triangles ABC and A1B1H1 are equilateral, they are equal on three sides, which was required to prove.