Question

Let f,g and h be the lengths of the perpendiculars from the circumcenter of $\Delta ABC$ on the sides a,b and c respectively. Prove that $\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=\frac{1}{4}\frac{abc}{fgh}$

Answer 1

Distance of circumcenter from vertex $A$ to side $BC$ is $R\cos A=f$
Similarly,

$g=R\cos B=f$; $h=R\cos C=f$
$\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=\frac{2R\sin A}{R\cos A}+\frac{2R\sin B}{R\cos B}+\frac{2R\sin C}{R\cos C}=2(\tan A+\tan B+\tan C)$

Also, $\frac{a}{f}.\frac{b}{g}.\frac{c}{h}=8\tan A\tan B\tan C$

But in triangle,

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

Thus,$\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=\frac{1}{4}\frac{abc}{fgh}$