Question

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.

Answer 1

Given:
$△ABC$ is an Equilateral Triangle.
$AB=BC=AC$
$CD$ Perpendicular to $AB$
To Prove:
$3A{C}^2=4C{D}^2$
Proof:

In $△ADC$

By Pythagoras Theorem,

$A{C}^2=A{D}^2+D{C}^2.....\left(1\right)$

In $△BDC$

By Pythagoras Theorem,

$B{C}^2=D{C}^2+B{D}^2.....\left(2\right)$

Adding 1) and 2) We get,

$2A{C}^2=2D{C}^2+A{D}^2+B{D}^2$ $(SinceAC=BC)$

Since, $AD=BD=\frac{1}{2}AC$ $(SinceAC=AB)$

$\therefore$ $2A{C}^2=2D{C}^2+\frac{1}{2}A{C}^2$

$\therefore$ $2A{C}^2-\frac{1}{2}A{C}^2=2D{C}^2$

$\therefore$ $3A{C}^2=4C{D}^2$