Let f, g and h be the lengths of the perpendicular from the circumcentre of $△ A B C$ on the sides a, b, and c,respectively.Prove that $\frac{a}{f} + \frac{b}{g} + \frac{c}{h} = \frac{1}{4} \frac{a b c}{f g h} .$

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Solution

Distance of circumcentre from to side BC is R cos A=f
Similarly, g=R cos B, h=R cos C
$\Rightarrow \frac{a}{f} + \frac{b}{g} + \frac{c}{h} = \frac{2 R s i n B}{R c o s A} + \frac{2 R s i n B}{R c o s B} + \frac{2 R s i n C}{R c o s C}$
$= 2 (t a n A + t a n B + t a n C)$
Also, $\frac{a}{f} \frac{b}{g} \frac{c}{h} = 8 t a n A t a n B t a n C$
But in triangle, $t a n A + t a n B t a n C = t a n A t a n B t a n C$ .
Thus $\frac{a}{f} + \frac{b}{g} + \frac{c}{h} = \frac{1}{4} \frac{a b c}{f g h}$

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Let f, g and h be the lengths of the perpendicular from the circumcentre of △ABC on the sides a, b, and c,respectively.Prove that af+bg+ch=14abcfgh.

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Let f, g and h be the lengths of the perpendicular from the circumcentre of $△ A B C$ on the sides a, b, and c,respectively.Prove that $\frac{a}{f} + \frac{b}{g} + \frac{c}{h} = \frac{1}{4} \frac{a b c}{f g h} .$