Question

If $x,y,z$ are the distance of the vertices of $△ABC$ respectively from the orthocentre, then prove that $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{abc}{xyz}$.

Answer 1

From the right$△AHF,$

$\cos \left(\angle HAF\right)=\frac{AF}{AH}-\left(1\right)$

$\therefore \cos \left(\angle HAF\right)=\cos (90-C)=\sin \left(\angle C\right)-\left(2\right)$

$\therefore \sin \left(\angle C\right)=\frac{AF}{AH}$

$⟹AH=\frac{AF}{\sin C}-\left(3\right)$

$Fromright△ABF,\cos \left(\angle A\right)=\frac{AF}{AB}$

$⟹AF=AB\times \cos A=2R.\sin C\cos A$

$\because Bysinlawof△le,\frac{AB}{\sin C}=2R,$

where R is the radius of the circle circumscribly, $the△ABC.$

$⟹AH=2R.\sin C.\frac{\cos A}{\sin C}=2R\cos A$

Thus,distance of the vertice$A$ from its orthocenter$=2R\cos A.$

Similarily, other 2 vertices are $2R\cos B\&2R\cos C$

$\therefore x=2R\cos A;y=2R\cos B;z=2R\cos C$

By sine law of$△ABC$

$a=2R\sin A;b=2R\sin B;c=2R\sin C$

$\therefore \frac{a}{x}=\tan A;\frac{b}{y}=\tan B;\frac{c}{z}=\tan C-\left(4\right)$

$\therefore \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\tan A+\tan B+\tan C-\left(5\right)$

$\therefore In△ABC,A+B+C={180}^{\circ }$

(Angle sum property of$△le)$

$⟹A+B=180-C\therefore \tan (A+B)=\tan (180-C)=-\tan C$

$\therefore \frac{\tan A+\tan B}{1-\tan A.\tan B}=-\tan C$

$\therefore \tan A+\tan B+\tan C=\tan A.\tan B.\tan C-\left(6\right)$

$\therefore \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\tan A.\tan B.\tan C-\left(7\right)$

Substituting the values of $\tan A,\tan B\&\tan Cfrom\left(4\right)$

$⟹\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{abc}{xyz}$