Prove that the ratio of the …

Given: ​{tex}\triangle {/tex}​ABC ​{tex} \sim {/tex}​ {tex}\triangle {/tex}DEF
AM is a median in ​{tex}\triangle {/tex}​ABC and DN is the corresponding median in {tex}\triangle {/tex}DEF
To prove: {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}
Proof: ​{tex}\triangle {/tex}​DBC ​{tex} \sim {/tex}​ {tex}\triangle {/tex}DEF
​{tex}\Rightarrow {/tex}​ ​{tex}\angle{/tex}​ A = ​{tex}\angle{/tex}​ D, ​{tex}\angle{/tex}​ B = ​{tex}\angle{/tex}​ E, ​{tex}\angle{/tex}​ C = ​{tex}\angle{/tex}​ F and

{tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}}{/tex}
Also, {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}}{/tex}={tex}\frac{{A{B^2}}}{{D{E^2}}} = \frac{{B{C^2}}}{{E{F^2}}} = \frac{{A{C^2}}}{{D{F^2}}}{/tex}........(i)[area theorem]
Now,{tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{\frac{1}{2}BC}}{{\frac{1}{2}EF}} = \frac{{BM}}{{EN}}{/tex}..............(ii)
In {tex}\triangle {/tex}ABM and DEN
​{tex}\angle{/tex}​ B= ​{tex}\angle{/tex}​ E and {tex}\frac{{AB}}{{DE}} = \frac{{BM}}{{EN}}{/tex} [From (ii)]
​{tex}\Rightarrow {/tex}​ ​{tex}\triangle {/tex}ABM ​{tex} \sim {/tex}​{tex}\triangle {/tex}DEN [SAS similarity]
​{tex}\Rightarrow {/tex}​{tex}\frac{{area\vartriangle ABM}}{{area\vartriangle DEN}} = \frac{{A{B^2}}}{{D{E^2}}} = \frac{{A{M^2}}}{{D{N^2}}} = \frac{{B{M^2}}}{{E{N^2}}}{/tex}............(iii)
From (i) and (iii), we get
{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}