If H is the orthocentre of triangle ABC and A | vedclass

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(C) Given that $H$ is the orthocentre of $	riangle ABC$ and $AH=x, BH=y, CH=z$.In any triangle,the distance from the orthocentre to the vertices is g

Vedclass · Accepted Answer

$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$

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(C) Given that $H$ is the orthocentre of $	riangle ABC$ and $AH=x, BH=y, CH=z$.In any triangle,the distance from the orthocentre to the vertices is given by $AH=2R \cos A$,$BH=2R \cos B$,and $CH=2R \cos C$,where $R$ is the circumradius.Thus,$x=2R \cos A, y=2R \cos B, z=2R \cos C$.We know that $a=2R \sin A, b=2R \sin B, c=2R \sin C$.Therefore,$\frac{a}{x} = \frac{2R \sin A}{2R \cos A} = 	an A$.Similarly,$\frac{b}{y} = 	an B$ and $\frac{c}{z} = 	an C$.We are looking for the expression $\frac{abc}{xyz} = \frac{a}{x} \cdot \frac{b}{y} \cdot \frac{c}{z} = 	an A 	an B 	an C$.In any triangle,$	an A + 	an B + 	an C = 	an A 	an B 	an C$.Thus,$\frac{abc}{xyz} = 	an A + 	an B + 	an C = \frac{a}{x} + \frac{b}{y} + \frac{c}{z}$.

If $H$ is the orthocentre of $\triangle ABC$ and $AH=x, BH=y, CH=z$,then $\frac{abc}{xyz}=$

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