Question

In $\Delta ABC,AD\perp BC$ and $BD:CD=3:1$. Prove that $2(A{B}^2-A{C}^2)=B{C}^2$

Answer 1

Given that:

In $△ABC,AD\perp BC,BD:CD=3:1$

To prove:

$2(A{B}^2-A{C}^2)=B{C}^2$

Solution:

$\frac{BD}{CD}=\frac{3}{1}$

or, $BD=3CD$

$BC=BD+CD$

or, $BC=3CD+CD=4CD$

In $△ADC$

By Pythagorean theorem

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

In $△ADB$

By Pythagorean theorem

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

Subtract (i) from (ii) we get,

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

$=(3CD{)}^2-C{D}^2$

$=9C{D}^2-C{D}^2$

$=8C{D}^2$

$=8{\left(\frac{BC}{4}\right)}^2$

$=8\times \frac{B{C}^2}{16}$

$=\frac{B{C}^2}{2}$

or, $2(A{B}^2-A{C}^2)=B{C}^2$ $\left[henceproved\right]$