Question

If $d_1,d_2,d_3$ are the diameters of three ex-circles of a $△ABC$, then $d_1{d}_2+d_2{d}_3+d_3{d}_1=$

• A. $(a+b+c{)}^2$
• B. $ab+bc+ca$
• C. $4{\Delta }^2$
• D. $2s^2$

Answer 1

Answer: (A)

We have,
$d_1=2r_1,d_2=2r_2,d_3=2r_3$
Now,
$d_1{d}_2+d_2{d}_3+d_3{d}_1=\left(2r_1\right)\left(2r_2\right)+\left(2r_2\right)\left(2r_3\right)+\left(2r_3\right)\left(2r_1\right)$
$=4\left[r_1{r}_2+r_2{r}_3+r_3{r}_1\right]$
$=4\left[\frac{\Delta }{s-a}\times \frac{\Delta }{s-b}+\frac{\Delta }{s-b}\times \frac{\Delta }{s-c}+\frac{\Delta }{s-c}\times \frac{\Delta }{s-a}\right]$
$=4\left[\frac{{\Delta }^2}{(s-a)(s-b)}+\frac{{\Delta }^2}{(s-b)(s-c)}+\frac{{\Delta }^2}{(s-c)(s-a)}\right]$
$=4{\Delta }^2\left[\frac{s-c+s-a+s-b}{(s-a)(s-b)(s-c)}\right]$
$=4{\Delta }^2\left[\frac{3s-(a+b+c)}{(s-a)(s-b)(s-c)}\right]$
$=4{\Delta }^2\left[\frac{3/2(a+b+c)-(a+b+c)}{(s-a)(s-b)(s-c)}\right]$
$=\frac{4{\Delta }^2}{2}\left[\frac{a+b+c}{(s-a)(s-b)(s-c)}\right]$
$=\frac{2s(s-a)(s-b)(s-c)(a+b+c)}{(s-a)(s-b)(s-c)}$
$=(a+b+c)\cdot (a+b+c)$
$=(a+b+c{)}^2$