I'm not sure what is meant by "proven algevraically", but I'll give an attempt --
Place the original triangle XYZ on a graph with the coordinates of
X = (0,0) Y = (a,0) Z = (b,c)
then the midpoint of XY = (a/2,0)
The similar triangle is X'Y'Z' with a similarity factor of k. Its coordinates will be
X' = (0,0) Y' = (ka,0) Z' = (kb,kc)
then the corresponding midpoint will be (ka/2,0)
The length of the median of the first triangle is
length = sqrt( (b - a/2)2 + (c - 0)2 ) = sqrt( (b - a/2)2 + c2 )
The length of the median of the second triangle is
length' = sqrt( (kb - ka/2)2 + (kc - 0)2 )
= sqrt( ( k(b - a/2) )2 + (kc)2 )
= sqrt( k2(b - a/2)2 + k2c2 )
= sqrt( k2 ) · sqrt( (b - a/2)2 + c2 )
= k·sqrt( (b - a/2)2 + c2 )
(which is k times the length of the median of the first triangle and so the medians have the same similarity
factor as the sides)