Question

prove that r1r2+r2r3+r3r1=s²=(a+b+c/2)²​

Answer 1

Answer:

Solution :

We know that,

i) r1 = A/( s-a ),

ii) r2 = A / (s - b),

iii) r3 = A/s - c)

LHS = 4(r1r2 + r2r3+r3r1)

= 4[A/(s-a)*A/(s-b) + A/(s-b)*A/(s-c)+A/ (s-c)*A/(s-a)]

=

4A² [(s-c+s-a+s-b)/(s-a)(s-b)(s-c)]

= 4A²[ {3s - (a+b+c)}/[(s-a)(s-b)(s-c)]

= 4A² [3s - 2s ]/[(s-a)(s-b)(s-c)]

[Since, a + b + c = 2s]

= 4A³ [ s/[(s-a)(s-b)(s-c)]

= 4A² {s²/[ s(s-a)(s-b)(s-c) ]}

4s² =

= ( 2s)²

= (a + b + c)²

= RHS