$A B C$ is a right triangle, right-angled at $C$ . Let $B C = a, C A = b, A B = c$ and let $p$ be the length of perepndicular from $C$ on $A B$ , prove that
i) $c p = a b$
ii) $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

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Solution

$G i v e n : △ A B C i s a r i g h t t r i a n g l e, r i g h t - a n g l e d a t C . L e t B C = a, C A = b, A B = c T o P r o v e : c p = a b \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} P r o o f : L e t C D ⊥ A B . T h e n, C D = p △ A r e a o f △ A B C = \frac{1}{2} (B a s e \times h e i g h t) = \frac{1}{2} (A B \times C D) = \frac{1}{2} c p A l s o, A r e a o f △ A B C = \frac{1}{2} (B C \times A C) = \frac{1}{2} a b ∴ \frac{1}{2} c p = \frac{1}{2} a b \Rightarrow c p = a b S i n c e △ A B C i s a r i g h t t r i a n g l e, r i g h t a n g l e d a t C . ∴ {A B}^{2} = {B C}^{2} + {A C}^{2} \Rightarrow c^{2} = a^{2} + b^{2} \Rightarrow ({\frac{a b}{p})}^{2} = a^{2} + b^{2} [∵ c p = a b \Rightarrow c = \frac{a b}{p}] \Rightarrow \frac{a^{2} b^{2}}{p^{2}} = a^{2} + b^{2} \Rightarrow \frac{1}{p^{2}} = \frac{1}{b^{2}} + \frac{1}{a^{2}} \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} H e n c e P r o v e d$

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