If in $Δ A B C$ the distances of the vertices from the orthocenter are x,y,and z,then prove that $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \frac{a b c}{x y z}$

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Solution

We know that distance of orthocenter (H) from vertex (A) is $2 R c o s A$
or
$x = 2 R cos A, y = 2 R cos B, z = 2 R cos C$

$\Rightarrow \frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \frac{2 R sin A}{2 R cos A} + \frac{2 R sin B}{2 R cos B} + \frac{2 R sin C}{2 R cos C}$

$= tan A + tan B + tan C = tan A tan B tan C$

Also, $\frac{a b c}{x y z} = \frac{(2 R sin A) (2 R sin B) (2 R sin C)}{(2 R cos A) (2 R cos B) (2 R cos C)} = tan A tan B tan C$

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