Question

In $△ABC$, show that $\frac{acosA+bcosB+ccosC}{a+b+c}=\frac{r}{R}$

Answer 1

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

So, $\frac{a\cos A+b\cos B+c\cos C}{a+b+c}$

$=R\left(\frac{\sin 2A+\sin 2B+\sin 2C}{a+b+c}\right)$

using $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$

so, we get $=4R\left(\frac{\sin A+\sin B+\sin C}{a+b+c}\right)$

$=\frac{abc}{2R^2\left(2S\right)}$

$=\frac{4R△}{4R^{2}S}$

so we get $\frac{△}{RS}=\frac{r}{R}$

$\therefore \frac{a\cos A+b\cos B+c\cos C}{a+b+c}=\frac{r}{R}$

Hence, proved.