Question

In given figure prove that three times the square of any side of an equilateral-triangle is equal to four times the square of the altitude.

Answer 1

In ABC an equilateral triangle, AD perpendicular to BC

In $\Delta ADB$ and $\Delta ADC$, we have

$AB=AC$ [Given]

$\angle B=\angle C$ [Each equal to ${60}^{\circ }$] and,

$\angle ADB=\angle ADC$ [Each equal to ${90}^{\circ }$]

$\therefore$ $\Delta ADB\cong \Delta ADC$

$\Rightarrow$ $BD=DC$

$\Rightarrow$ $BD=DC=\frac{1}{2}BC$

Since $\Delta ADB$ is a right-angled triangle at D.

$\therefore$ $A{B}^2=A{D}^2+B{D}^2$

$\Rightarrow$ $A{B}^2=A{D}^2+(\frac{1}{2}BC{)}^2$

$\Rightarrow$ $A{B}^2=A{D}^2+\frac{B{C}^2}{4}$

$\Rightarrow$ $A{B}^2=A{D}^2+\frac{A{B}^2}{4}$ $[\because BC=AB]$

$\Rightarrow$ $\frac{3}{4}A{B}^2=A{D}^2$

$\Rightarrow$ $3A{B}^2=4A{D}^2$ $\left[Henceproved\right]$