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(N/A) Let the side of the equilateral triangle be $a$,and $AE$ be the altitude of $	riangle ABC$.Since the altitude of an equilateral triangle bisect

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(N/A) Let the side of the equilateral triangle be $a$,and $AE$ be the altitude of $\triangle ABC$.
Since the altitude of an equilateral triangle bisects the base,
$BE = EC = \frac{BC}{2} = \frac{a}{2}$
Applying the Pythagoras theorem in the right-angled $\triangle ABE$,we obtain:
$AB^2 = AE^2 + BE^2$
Substituting the values $AB = a$ and $BE = \frac{a}{2}$:
$a^2 = AE^2 + (\frac{a}{2})^2$
$a^2 = AE^2 + \frac{a^2}{4}$
Rearranging the terms to solve for $AE^2$:
$AE^2 = a^2 - \frac{a^2}{4}$
$AE^2 = \frac{4a^2 - a^2}{4}$
$AE^2 = \frac{3a^2}{4}$
Multiplying both sides by $4$:
$4AE^2 = 3a^2$
Thus,it is proved that $4 \times$ (square of altitude) $= 3 \times$ (square of one side).

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.

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