Question

Let $f,g,h$ be the lengths of the perpendiculars from the circumcenter of the $\Delta ABC$ on the BC, CA, and AB, respectively, if $\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=\lambda \frac{abc}{fgh},$ then the value of $"\lambda "$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{1}{4}$

We know that angle made at the circumcenter is double the angle of the remaining vertex.

In $\Delta OBD;\tan A=\frac{a}{2f}⟹\frac{a}{f}=2\tan A$

Similarly; $\tan B=\frac{b}{2g}⟹\frac{b}{g}=2\tan B$

$\tan C=\frac{c}{2h}⟹\frac{c}{h}=2\tan C$

Now;

$\frac{a}{f}+\frac{b}{c}+\frac{c}{h}=2\tan A+2\tan B+2\tan C$

$=2\tan A.\tan B.\tan C.$ $[\text{If}A+B+C=180°,\tan A+\tan B+\tan C=\tan A.\tan B.\tan C]$

$2\times \frac{a}{2f}\times \frac{b}{2g}\times \frac{c}{2h}$

$=\frac{1}{4}\frac{abc}{fgh}$

Hence;

$\lambda =1/4$