In $△ A B C$ , $∠ A B C = 90^{o}$ , $B D ⊥ A C$ . If $A B =$ ' $c$ ' units, $B C =$ ' $a$ ' units, $B D =$ ' $p$ ' units, $C A =$ ' $b$ ' units.
Prove that $\frac{1}{a^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$ .

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Solution

Given $∠ A B C = 90^{o}$ , $B D$ is perpendicular to $A C, A B = c; B C = a; B D = p; C A = b$ .
Area of $△ A B C = \frac{1}{2} \times A C \times B D$
$= \frac{1}{2} \times b \times p = \frac{1}{2} b p$
Area of $△ A B C = \frac{1}{2} \times B C \times A B$
$= \frac{1}{2} c a$
$\Rightarrow \frac{1}{2} b p = \frac{1}{2} c a$
$∴ b p = c a$
$\Rightarrow b = \frac{c a}{p}$
In right triangle $A B C$ ,
$A C^{2} = A B^{2} + B C^{2}$
$\Rightarrow b^{2} = c^{2} + a^{2}$
$∴ (\frac{c a}{p})^{2} = c^{2} + a^{2}$
$\Rightarrow \frac{c^{2} a^{2}}{p^{2}} = c^{2} + a^{2}$
$∴ \frac{1}{p^{2}} = \frac{c^{2} + a^{2}}{c^{2} a^{2}}$
$∴ \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{c^{2}}$

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