Question

In given figure the perpendicular $AD$ on the base $BC$ of a $△ABC$ intersects $BC$ at $D$ so that $DB=3CD$. Prove that $2{AB}^2=2{AC}^2+{BC}^2$.

Answer 1

We have,

$DB=3CD$

$\therefore$ $BC=BD+DC$

$\Rightarrow$ $BC=3CD+CD$

$\Rightarrow$ $BC=4CD$

$\Rightarrow$ $CD=\frac{1}{4}BC$

$\therefore$ $CD=\frac{1}{4}BC$ and $BD=3CD=\frac{3}{4}BC$.......(i)

Since $△ABD$ is a right-angled triangle at D

$\therefore$ ${AB}^2={AD}^2+{BD}^2$......(ii)

Similarly, $△ACD$ is a right-angled triangle at D.

$\therefore$ ${AC}^2={AD}^2+{CD}^2$.......(iii)

Subtracting equation (iii) from equation (ii) we get

${AB}^2-{AC}^2={BD}^2-{CD}^2$

$\Rightarrow$ ${AB}^2-{AC}^2={\left(\frac{3}{4}BC\right)}^2-{\left(\frac{1}{4}BC\right)}^2$ [From (i) $CD=\frac{1}{4}BC,BD=\frac{3}{4}BC$]

$\Rightarrow$ ${AB}^2-{AC}^2=\frac{9}{16}{BC}^2-\frac{1}{16}{BC}^2$

$\Rightarrow$ ${AB}^2-{AC}^2=\frac{1}{2}{BC}^2$

$\Rightarrow$ $2({AB}^2-{AC}^2)={BC}^2$

$\Rightarrow$ $2{AB}^2=2{AC}^2+{BC}^2$ $\left[Henceproved\right]$