Prove that in a right angel …
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Sia ? 6 years, 4 months ago
Given : {tex}\triangle \mathrm{ABC}{/tex} is right angle at B
To prove: {tex}A C^{2}=A B^{2}+B C^{2}{/tex}
Construction: Draw {tex}B D \perp A C{/tex}
Proof:In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}ABD
{tex}\angle{/tex}ABC={tex}\angle{/tex}ADB=90
{tex}\angle{/tex}A is common
{tex}\angle{/tex}ABC{tex}\sim{/tex}{tex}\angle{/tex}ADB
{tex}\Rightarrow \frac{A D}{A B}=\frac{A B}{A C}{/tex}
{tex}\Rightarrow{/tex} AD. AC = AB2 ...(i)
Similarly {tex}\Delta \mathrm{BDC} \sim \Delta \mathrm{ABC}{/tex}
{tex}\Rightarrow \frac{C D}{B C}=\frac{B C}{A C}{/tex}
{tex}\Rightarrow{/tex} CD.AC = BC2 ...(ii)
Adding (i) and (ii)
AD.AC + CD. AC= AB2 + BC2
AC(AD + CD) = AB2 + BC2
AC {tex}\times{/tex} AC = AB2 + BC2
AC2 = AB2 + BC2
Hence Proved
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