Mark the Correct alternative in the following:
In any ∆ABC,
a2(sin B – sin C) + b2(sin C – sin A) + c2(sin A – sin B)
a2sinB – a2sinC + b2sinC – b2sinA + c2sinA – c2sinB
a2sinB – c2sinB + b2sinC – a2sinC + c2sinA – b2sinA
(a2 – c2)sin B + (b2 – a2)sin C + (c2 – b2)sin A
By Sine law,
(a2 – c2)bk + (b2 – a2)ck + (c2 – b2)ak
a2bk – c2bk + b2ck – a2ck + c2ak – b2ak
Considering it as equilateral,
a=b=c
a2bk – c2bk + b2ck – a2ck + c2ak – b2ak=0
Option D