Question

If $d_{1}, d_{2}, d_{3}$ are the diameters of three ex-circles of a $\triangle ABC$, then $d_{1}\, d_{2}+d_{2} \,d_{3}+d_{3} \,d_{1}=$

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

Answer 1

Answer: (A)

We have,
$d_{1}=2 r_{1}, d_{2}=2 r_{2}, d_{3}=2 r_{3}$
Now,
$d_{1} d_{2}+d_{2} d_{3}+d_{3} d_{1}=\left(2 r_{1}\right)\left(2 r_{2}\right)+\left(2 r_{2}\right)\left(2 r_{3}\right)+\left(2 r_{3}\right)\left(2 r_{1}\right)$
$=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{3 s-(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] $
$=\frac{4 \Delta^{2}}{2}\left[\frac{a+b+c}{(s-a)(s-b)(s-c)}\right]$
$=\frac{2 s(s-a)(s-b)(s-c)(a+b+c)}{(s-a)(s-b)(s-c)}$
$=(a+b+c) \cdot(a+b+c)$
$=(a+b+c)^{2}$