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# In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of $$\triangle$$ ABC.
BE = EC = $$\frac{BC}{2} = \frac{a}{2}$$
In $$\triangle$$ ABE, by Pythagoras Theorem, we get
$$\Rightarrow AB^2 = AE^2 + BE^2$$
$$\Rightarrow a^2 = AE^2 + ({{a} \over {2}})^2$$
$$\Rightarrow AE^2 = a^2 - {{a^2} \over {4}}$$
$$\Rightarrow AE^2 = {{3a^2} \over {4}}$$
$$\Rightarrow 4AE^2 = 3a^2$$

4 × (Square of altitude) = 3 × (Square of one side)
Hence, proved.