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

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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