Proof that three times the sum …

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Sia ? 6 years, 6 months ago
Given: {tex}\triangle {/tex} ABC in which AB=BC=CA and AD {tex} \bot {/tex} BC

To prove: 3BC2 = 4AD2
Proof: {tex}\triangle {/tex} ABC is an equilateral triangle. AD {tex} \bot {/tex} BC
{tex}\Rightarrow {/tex} BD = DC [In an equilateral,{tex}\triangle {/tex} altitude bisect the opposite side]
Now in {tex}\triangle {/tex}ADB, {tex}\angle{/tex} D = 90o
{tex}\Rightarrow {/tex} AB2 = AD2 + BD2 [Pythagoras theorem]
{tex}\Rightarrow {/tex} BC2 = AD2+{tex}{\left( {\frac{{BC}}{2}} \right)^2}{/tex}
{tex}\Rightarrow {/tex} 4BC2 = 4AD2 + BC2
{tex}\Rightarrow {/tex} 3BC2 = 4AD2
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