Prove that ratio of area of …

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Prove that ratio of area of two similar triangle is equal to ratio of square of angle bisector

Posted by Kushagra Jaiswal 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given: {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}DEF
AM is the bisector of {tex}\angle{/tex} BAC and is the corresponding bisector of {tex}\angle{/tex} EDF
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}.......(i) [Area theorem]
Now {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}DEF
{tex}\Rightarrow {/tex} {tex}\angle{/tex} A = {tex}\angle{/tex} D {tex}\Rightarrow {/tex} {tex}\frac{1}{2}\angle A = \frac{1}{2}\angle D{/tex}
{tex}\Rightarrow {/tex} {tex}\angle{/tex}BAM = {tex}\angle{/tex}EDN
In {tex}\triangle {/tex}ABM and DEN {tex}\angle{/tex} B = {tex}\angle{/tex} E and {tex}\angle{/tex} BAM= {tex}\angle{/tex}EDN
{tex}\Rightarrow {/tex} {tex}\triangle {/tex}ABM {tex} \sim {/tex} {tex}\triangle {/tex}DEN
{tex}\Rightarrow {/tex} {tex}\frac{{AB}}{{DE}} = \frac{{AM}}{{DN}} \Rightarrow \frac{{A{B^2}}}{{D{E^2}}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}...................(ii)
{tex}\Rightarrow {/tex} {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex} [From (i) and (ii)] Proved.

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CBSE > Class 10 > Mathematics