Problem Statement
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.
Verified Solution & Marking Scheme
Draw Altitude in Equilateral Triangle
Let $\triangle ABC$ be an equilateral triangle with side $a$. Draw altitude $AD \perp BC$.
In an equilateral triangle, the altitude bisects the base: $BD = DC = \frac{a}{2}$.
Apply Pythagoras Theorem in Right Triangle ABD
$AB^2 = AD^2 + BD^2$
$a^2 = AD^2 + \left(\frac{a}{2}\right)^2 = AD^2 + \frac{a^2}{4}$
$a^2 - \frac{a^2}{4} = AD^2 \implies \frac{3a^2}{4} = AD^2$
Clear Denominator
$3a^2 = 4 AD^2 \implies 3 AB^2 = 4 AD^2$