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Theorem of Geometric Mean

In ABC, ∠ C...

Question

In $△ A B C, ∠ C = 90^{o}$ if $p$ is length of in from $C$ to $A B$ . $A B = c, B C = a, C A = b$ show that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

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Solution

Given altitude be $p$
In $△ A B C, ∠ C = 90^{\circ} ⟹ c^{2} = a^{2} + b^{2}$
area of $△ A B C = \frac{1}{2} c p$
Area of $△ A B C = \frac{1}{2} a b (∵$ right angled triangle $)$
$a b = c p$
$a^{2} b^{2} = c^{2} p^{2}$
$\frac{1}{p^{2}} = \frac{a^{2} + b^{2}}{a^{2} b^{2}}$
$\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

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