Question

In given figure $\Delta ABC$ is a triangle where $\angle C={90}^o$, let $BC=a,CA=b,AB=c$ and let '$p$' be the length of the perpendicular from $C$ on $AB$. Prove that:
(i) $cp=ab$ (ii) $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

Answer 1

Let $CD\perp AB$. then, $CD=p$.

$\therefore$ Area of $△ABC=\frac{1}{2}(Base\times Height)$

$\Rightarrow$ Area of $△ABC=\frac{1}{2}(AB\times CD)=\frac{1}{2}cp$

Also,

Area of $△ABC=\frac{1}{2}(BC\times AC)=\frac{1}{2}ab$

$\therefore$ $\frac{1}{2}cp=\frac{1}{2}ab$

$\Rightarrow$ $cp=ab$ $\left[Henceproved\right]$

(ii) Since $△ABC$ is a right-angled triangle at C.

$\therefore$ ${AB}^2={BC}^2+{AC}^2$

$\Rightarrow$ $c^2=a^2+b^2$

$\Rightarrow$ ${\left(\frac{ab}{p}\right)}^2=a^2+b^2$ [$\because cp=ab\therefore c=\frac{ab}{p}$]

$\Rightarrow$ $\frac{a^2{b}^2}{p^2}=a^2+b^2$

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

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

$\Rightarrow$ $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$ $\left[Henceproved\right]$