Question

在三角形ABC中,求证(a^2-b^2)/(sinA+sinB)+(b^2-c^2)/(sinB+sinC)+(c^2-

Answer 1

用正弦定理

SinA=a/2r

示例一个：(a^2-b^2)/(sinA+sinB)

=[(a+b)(a-b)]/[(a+b)/2r]

=2r(a-b)

同理以此类推

(a^2-b^2)/(sinA+sinB)+(b^2-c^2)/(sinB+sinC)+(c^2-a^2)/(sinC+sinA)

=2r(a-b)+2r(b-c)+2r(c-a)

=0