Question

In the given figure, $AD\perp BC$ and $BD=\frac{1}{3}CD$ prove that $2A{C}^2=2A{B}^2+B{C}^2.$

Answer 1

In right $\Delta ABD,$ by pythogoras theorem.

$\therefore A{B}^2=A{D}^2+B{D}^2$

In right $\Delta ADC,$ by pythagoras theorem,

$\therefore A{C}^2=A{D}^2+D{C}^2$

Subtracting (ii) from (i) we get,

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

$={\left[\frac{1}{3}CD\right]}^2-D{C}^2$

$=\frac{C{D}^2}{9}-\frac{D{C}^2}{1}=\frac{-8C{D}^2}{9}$

Since, $\frac{BD}{CD}=\frac{1}{3}\Rightarrow \frac{BD}{CD}+1=\frac{1}{3}+1$

$\Rightarrow \frac{BD+CD}{CD}=\frac{4}{3}\Rightarrow \frac{BC}{CD}=\frac{4}{3}$

$\Rightarrow CD=\frac{3}{4}BC.$

$\therefore A{B}^2-A{C}^2=\frac{-8}{9}{\left[\frac{3BC}{4}\right]}^2$

$=\frac{-8}{9}\times \frac{9}{16}B{C}^2=-\frac{1}{2}B{C}^2$