Prove that the areas of two …
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Sia ? 6 years, 5 months ago
Given: {tex}\triangle {/tex} ABC {tex} \sim {/tex}{tex}\triangle {/tex}DEF
AM {tex} \bot {/tex} BC and DN {tex} \bot {/tex} EF
To prove:{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}
proof::{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{B^2}}}{{D{E^2}}}{/tex}................[Area theorem]
In {tex}\triangle {/tex} ABM and DEN
{tex}\angle{/tex} ABM and {tex}\angle{/tex} DEN [ {tex}\triangle {/tex} ABC {tex} \sim {/tex} {tex}\triangle {/tex}DEF]
{tex}\angle{/tex}AMB = {tex}\angle{/tex}DNE [90o each]
{tex}\Rightarrow {/tex} {tex}\triangle {/tex}ABM {tex} \sim {/tex}{tex}\triangle {/tex}DEN [AA similarity]
{tex}\therefore {/tex} {tex}\frac{{AB}}{{DE}} = \frac{{AM}}{{DN}}{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{{A{B^2}}}{{D{E^2}}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex} .....(ii)
From (i) and (ii) we get
{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}.
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