Question

$ABC$ is a right angled triangle with $\angle C={90}^o,BC=a,AC=b,CD\perp AB$ and $CD=P$. Show that $\frac{1}{P^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

Given that:

$\angle C={90}^{\circ },BC=a,AC=b,CD\perp AB,CD=P$

To Prove:

$\frac{1}{P^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Solution:

$A\left(ABC\right)=\frac{1}{2}ab=\frac{1}{2}P\times AB$

or, $AB=\frac{ab}{P}$

In right angled $△ACB,$

By Pythagorean theorem,

$A{B}^2=B{C}^2+A{C}^2$

or, ${\left(\frac{ab}{P}\right)}^2=a^2+b^2$

or, $\frac{a^2{b}^2}{P^2}=a^2+b^2$

Dividing both sides by $a^2{b}^2$ we get,

or, $\frac{1}{P^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Hence, proved.