Prove that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes.
Let $A B C$ be an equilateral triangle and let $A D \perp B C$.
In $\triangle A D B$ and $\triangle A D C$ we have
$A B=A C$
$\angle B=\angle C$
And $\angle A D B=\angle A D C$
$\Rightarrow \triangle A D B \cong \triangle A D C$
So, $B D=D C$
$\Rightarrow B D=D C=\frac{1}{2} B C$
Since $\triangle A D B$ is a right triangle right-angled at $D .$ So
$A B^{2}=A D^{2}+B D^{2}$
$A B^{2}=A D^{2}+\left(\frac{1}{2} B C\right)^{2}$
$A B^{2}=A D^{2}+\frac{B C^{2}}{4}$
$A B^{2}=A D^{2}+\frac{A B^{2}}{4}$
$\frac{3}{4} A B^{2}=A D^{2}$
$3 A B^{2}=4 A D^{2}$
Hence proved.
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