Question

A point $D$ is on the side of an equilateral triangle $ABC,$ such that $DC=\frac{1}{4}BC$ then:

• A. $2A{D}^2=C{D}^2$
• B. $A{D}^2=3C{D}^2$
• C. $2A{D}^2=7C{D}^2$
• D. $A{D}^2=13C{D}^2$

Answer 1

The correct option is D $A{D}^2=13C{D}^2$

Given in, $△ABC$, $BD=\left(\frac{1}{4}\right)BC$

Let, $AB=BC=CA=a$

Hence, $DC=\left(\frac{3}{4}\right)BC=\frac{3a}{4}$

Draw $AE\perp BC$

$BE=EC=\left(\frac{1}{2}\right)BC=\frac{a}{2}$

Since $ABC$ is equilateral triangle,

Hence $DE=BE-BD$

In $△AED$, by Pythagoras theorem we have:

$A{D}^2=A{E}^2+D{E}^2$

$={\left(\frac{\sqrt{3}a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^4=\frac{3a^2}{4}+\frac{a^2}{16}$

$=\frac{13a^2}{16}$

$\therefore 16A{D}^2=13B{C}^2=13(4DC{)}^2$

$\Rightarrow 16A{D}^2=13\times 16D{C}^2$

$⟹A{D}^2=13D{C}^2$