Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Given: AM and DN are the medians of triangles ABC and DEF respectively and \(\triangle\) ABC ~ \(\triangle\) DEF.



We have to prove: Area(\(\triangle\) ABC)/Area(\(\triangle\) DEF) = AM2/DN2

Since, \(\triangle\) ABC ~ \(\triangle\) DEF (Given)
Area(\(\triangle\) ABC)/Area(\(\triangle\) DEF) = \((AB^2/DE^2)\) ……………………………(i)
and, AB/DE = BC/EF = CA/FD ………………………………………(ii)
\({{AB} \over {DE}} ={{{{1} \over {2}}BC} \over {{{1} \over {2}}EF}} = {{CD} \over {FD}} \)

In \(\triangle\) ABM and \(\triangle\) DEN,

Since \(\triangle\) ABC ~ \(\triangle\) DEF
\(\angle\) B = \(\angle\) E
AB/DE = BM/EN [Already Proved in equation (i)]
\(\triangle\) ABC ~ \(\triangle\) DEF [SAS similarity criterion]
AB/DE = AM/DN …………………………………………………..(iii)
\(\triangle\) ABM ~ \(\triangle\) DEN

As the areas of two similar triangles are proportional to the squares of the corresponding sides.
area(\(\triangle\) ABC)/area(\(\triangle\) DEF) = \(AB^2/DE^2 = AM^2/DN^2\)
Hence, proved.