Question

With respect to general notations, Show that

$\frac{r_1(r_2+r_3)}{\sqrt{r_1{r}_2+r_2{r}_3+r_3{r}_1}}=a$

Answer 1

$\frac{r_1(r_2+r_3)}{\sqrt{r_1{r}_2+r_2{r}_3+r_3{r}_1}}$

$r_1=\frac{\Delta }{r-a};r_2=\frac{\Delta }{r-b};r_3=\frac{\Delta }{r-c}$

G.E$=\frac{\frac{\Delta }{r-a}(\frac{\Delta }{r-b}+\frac{\Delta }{r-c})}{\sqrt{\frac{{\Delta }^2}{(r-a)(r-b)}+\frac{{\Delta }^2}{(r-b)(r-c)}+\frac{{\Delta }^2}{(r-c)(r-a)}}}$

$=\frac{\frac{{\Delta }^2}{r-a}\left[\frac{r-c+r-b}{(r-b)(r-c)}\right]}{\sqrt{r(r-c)+r(r-a)+r(r-b)}}$

$=\frac{\frac{{\Delta }^2}{(r-a)(r-b)(r-c)}[2r-b-c]}{\sqrt{3r^2-r(a+b+c)}}$

$=\frac{\frac{r(r-a)(r-b)(r-c)}{(r-a)(r-b)(r-c)}\cdot [a+b+c-b-c]}{\sqrt{3r^2-2r^2}}$

$=\frac{r\left(a\right)}{\sqrt{r^2}}=\frac{r\left(a\right)}{r}=a$.