# if A+B+C=Pi , ( a, b , c are sides of triangle) prove that(a^2)*[sin(B-C)]/(sinB+sinC) + (b^2)*sin(C-A)/(sinC+sinA) + (c^2)sin(A-B)/(sinA+sinB) = 0 at least say steps and what formula to use

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Let us consider first term of the given expression

`(a^2sin(B-C))/{sin(B)+sin(C)}`

`={a^2 2 sin((B-C)/2)cos((B-C)/2)}/{2sin((B+C)/2)cos((B-C)/2)}`

`B+C=pi-A`

`(B+C)/2=pi/2-A/2`

`sin((B+C)/2)=sin(pi/2-A/2)=cosA/2`

`a/(sin(A))=b/(sin(B))=c/(sin(c))=k`

`a=ksin(A)`

`a=2ksin(A/2)cos(A/2)`

`={a^2sin((B-C)/2)}/(cos(A/2))`

`={a(2ksin(A/2)cos(A/2))sin((B-C)/2)}/(cos(A/2))`

`=2ka sin(A/2)sin((B-C)/2)`

`=2kacos((B+C)/2)sin((B-C)/2)`

`=ka(sin(B)-sin(C))`

`Thus`

`{a^2sin(B-C)}/{sin(B)+sin(C)}=ka(sin(B)-sin(C))`

`so`

`LHS=``{a^2sin(B-C)}/{sin(B)+sin(C)}+{b^2sin(C-A)}/{sin(C)+sin(A)}+{c^2sin(A-B)}/{sin(A)+sin(B)}`

`=ka(sin(B)-sin(C))+kb(sin(C)-sin(A))+kc(sin(A)-sin(B))`

`=k{(a-c)sin(B)+(b-a)sin(C)+(c-b)sin(A)}`

`=k{k(sin(A)-sin(C))sin(B)+k(sin(B)-sin(A))sin(C)+k(sin(C)-sin(B))sin(A)}`

`=k^2{sin(A)sin(B)-sin(C)sin(B)+sin(B)sin(C)-sin(A)sin(C)+sin(C)sin(A)-sin(B)sin(A)}`

`=k^2{0}`

`=0`

`=RHS`

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