Question

$ABC$ is a right triangle right angled at $C.$ Let $BC=a,CA=b,AB=c$ and let $p$ be the length of perpendicular from $C$ on $AB$ prove that

i) $pc=ab$

ii) $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

In $△ABC,$ $CD$ perpendicular to $AB.$

(i) To Prove:- $pc=ab$

In $△ABC,$

Considering base $BC$ and height $AC,$

$ar\left(ABC\right)=\frac{1}{2}\times a\times b\dots \left(i\right)$

Considering base $AB$ and height $CD,$

$ar\left(ABC\right)=\frac{1}{2}\times p\times c\dots \left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right),$

$\begin{array}{cc}\frac{1}{2}\times p\times c& =\frac{1}{2}\times a\times b\\ pc& =ab\end{array}$

(ii) Now,

$\begin{array}{cc}pc& =ab\\ \frac{c}{ab}& =\frac{1}{p}\\ \frac{c^2}{a^2{b}^2}& =\frac{1}{p^2}\\ \frac{a^2+b^2}{a^2{b}^2}& =\frac{1}{p^2}\left[\text{By using Pythagoras Theorem}\right]\\ \frac{1}{a^2}+\frac{1}{b^2}& =\frac{1}{p^2}\end{array}$

Hence, proved.