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Preeti Dabral 1 year, 9 months ago
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}
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