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Theorem of Geometric Mean

Δ ABC is righ...

Question

$Δ A B C$ is right angled triangle . if p is the length of the perpendicular from C to AB and a.b.c are the lengths of sides opposite $∠ A, ∠ B, ∠ C,$ respectively then prove that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} .$

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Solution

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Solution : -

Area of $△ A B C = \frac{1}{2} b a$ __(1)

also P is the perpendicular distance

$∴$ Area $= \frac{1}{2} P C$ _(2)

From (1) & (2)

$a b = p c$ _(3)

but we know $C = \sqrt{a^{2} + b^{2}}$

$\Rightarrow C^{2} = a^{2} + b^{2}$

Now by (3)

$a^{2} b^{2} = P^{2} C^{2} = P^{2} (a^{2} + b^{2})$

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

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

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