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An equilateral triangle ABC prove that …

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An equilateral triangle ABC prove that 16 AD SQUARE = 13BC SQUARE

    • Posted by Arjun Kapoor 6 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 6 years, 6 months ago

    In equilateral {tex}\triangle{/tex}ABC. 4BD = BC
    Construction: Draw AE {tex}\perp{/tex} BC. {tex}\therefore{/tex} BE = {tex}\frac{1}{2}{/tex}BC.
    In right {tex}\triangle{/tex}AED, AD2 = DE2 + AE2 {tex}\Rightarrow{/tex} AE2 = AD2 - DE2 ..(i)
    In right {tex}\triangle{/tex}AEB, AB2 = AE2 + BE2 {tex}\Rightarrow{/tex} AB2 = AD2 - DE2 + BE2 [using (i)]

    {tex}\Rightarrow{/tex} AB2 + DE2 - BE2 = AD2
    {tex}\Rightarrow{/tex} AB2 + (BE - BD)2 - BE2 = AD2
    {tex}\Rightarrow{/tex} AB2 + BE2 + BD2 - 2BE.BD - BE2 = AD2
    {tex}\Rightarrow{/tex} AB2 + ({tex}\frac{1}{2}{/tex}BC)2 - {tex}2 \times \frac{1}{2}{/tex}BC{tex}\times \frac{1}{4}{/tex}BC = AD2
    {tex}\Rightarrow{/tex} AB2 + {tex}\frac{1}{16}{/tex}BC2 - {tex}\frac{1}{4}{/tex}BC2 = AD2
    {tex}\Rightarrow{/tex} BC2 - {tex}\frac{3}{16}{/tex}BC2 = AD2 [{tex}\because{/tex} AB = BC]
    {tex}\Rightarrow{/tex} {tex}\frac { 13 \mathrm { BC } ^ { 2 } } { 16 }{/tex} = AD2
    {tex}\Rightarrow{/tex} 13BC2 = 16AD2

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