Prove that for equal triangles ABC and A1B1C1 the medians drawn from the vertices A and A1 are equal.

Given: triangles ABC = A1B1C1;
AM median of triangle ABC;
A1M1 median of triangle A1B1C1;
Prove that AM = A1M1
Solution: Consider triangles АМС and А1М1С1:
1) AC = A1C1 as triangles ABC = A1B1C1;
2) angle С = angle С1 in the same way as triangles ABC = А1В1С1;
3) MC = M1C1 as AM median (the median divides the side to which it is drawn into two equal parts) and BM = MC and A1M1 median and B1M1 = M1C1;
4) triangles АМС = А1М1С1 according to the first sign of equality of triangles;
5) AM = A1M1.
Proven.