Question

In an equilateral $△ABC$, $AD\perp BC$. Prove that $A{B}^2+C{D}^2=\frac{5}{4}A{C}^2$

Answer 1

Given that:

$ABC$ is an equilateral triangle. $AD\perp BC$ i.e. $BD=DC$

To Prove:

$A{B}^2+C{D}^2=\frac{5}{4}A{C}^2$

Solution:

In $△ADB,$

By Pythagorean theorem,

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

In $△ADC,$

By Pythagorean theorem,

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

Subtract (ii) from (i) we get,

$A{B}^2-A{C}^2=B{D}^2-C{D}^2$

or, $A{B}^2+C{D}^2=B{D}^2+A{C}^2$

or, $A{B}^2+C{D}^2={\left(\frac{1}{2}BC\right)}^2+A{C}^2$ $\because BD=CD=\frac{1}{2}BC$

or, $A{B}^2+C{D}^2={\left(\frac{BC}{2}\right)}^2+A{C}^2$

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

or, $A{B}^2+C{D}^2=\frac{A{C}^2}{4}+A{C}^2$ $\because AB=BC=AC$

or, $A{B}^2+C{D}^2=\frac{5}{4}A{C}^2$

Hence, proved.