If two triangles are similar, then the lengths of corresponding medians are in the same ratio as the lengths of corresponding sides.

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)