Question

In a triangle $ABC,\angle A={90}^o$, and $AD$ is the altitude. Complete the relation $\frac{BD}{BA}=\frac{AB}{DB}-\left(\frac{.......}{AB\times BD}\right)$

Answer 1

$\frac{BD}{BA}=\frac{AB}{BD}-\frac{x}{AB\times BD}$

$\frac{BD}{BA}=\frac{A{B}^2-x}{\left(BD\right)\left(BA\right)}$

$\Rightarrow {BD}^2=A{B}^2-A{D}^2$

$\Rightarrow x=A{B}^2-B{D}^2$

$\Rightarrow x=A{D}^2$ (as in $△ABD$, $A{D}^2+B{D}^2=A{B}^2$)