Question

If $\Delta ABC$ is an equilateral triangle such that $AD$ is perpendicular to $BC,$ then $A{D}^2$ = ___.

• A. $\frac{3}{2}D{C}^2$
• B. $2D{C}^2$
• C. $3C{D}^2$
• D. $4D{C}^2$

Answer 1

The correct option is C

$3C{D}^2$

We first note that the triangles, $△ADC$ and $△ADB$ are congruent by RHS criterion of congruence.
($\because AB=AC..(△\text{ABC is equilateral)}$
$AD=AD..\left(\text{Common hypotenuse}\right)$
$\angle ADB=\angle ADC={90}^{\circ }$)
Now, using Pythagoras' theorem in $△ADC$ as $AD\perp BC,$ we have
$A{C}^2=A{D}^2+C{D}^2.$
$\Rightarrow A{D}^2=A{C}^2-C{D}^2$
$\Rightarrow A{D}^2=B{C}^2-C{D}^2(\because △\text{ABC is equilateral})$
$\Rightarrow A{D}^2=(2CD{)}^2-C{D}^2(\because \text{BD=CD by CPCT})$
$\Rightarrow A{D}^2=3C{D}^2$