Question

With usual notations, prove that in a triangle ABC
$\frac{b-c}{r_1}+\frac{c-a}{r_2}+\frac{a-b}{r_3}=0$

Answer 1

In any given triangle ABC, with usual notations, we have a relation

$r_1=\frac{\Delta }{s-a}$

$r_2=\frac{\Delta }{s-b}$

$r_3=\frac{\Delta }{s-c}$

where, $\Delta$ is Area of given triangle,

a,b,c are sides of triangle and s is semi- perimeter,

$r_1,r_2,r_3$ are the radii of ex-circles opposite to the vertices of A, B ,C of triangle ABC

Now, putting above relation of r1, r2 and r3 in the left side of given equality, we get

$\frac{b-c}{\left(\frac{\Delta }{s-a}\right)}+\frac{c-a}{\left(\frac{\Delta }{s-b}\right)}+\frac{a-b}{\left(\frac{\Delta }{s-c}\right)}=0$

On rearranging, we get

$\frac{(b-c)(s-a)}{\Delta }+\frac{(c-a)(s-b)}{\Delta }+\frac{(a-b)(s-c)}{\Delta }=0$ $=0$

Now multiplying both side by $\Delta$, equation reduces to

$(b-c)(s-a)+(c-a)(s-b)+(a-b)(s-c)=0$

on solving further by opening brackets,

$b\cdot s-b\cdot a-c\cdot s+c\cdot a+c\cdot s-c\cdot b-a\cdot s+a\cdot b+a\cdot s-a\cdot c-b\cdot s+b\cdot c=0$

In left side of equation each term gets cancelled and we obtain

$0=0$

hence proved.