Prove that the ratio of the …

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

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