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Must be proven algebraically. 

 Jan 24, 2022
edited by Guest  Jan 24, 2022
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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)

 Jan 24, 2022

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