Conjecture: In any triangle ABC,

a(sinB - sinC) + b(sinC - sinA) + c(sinA - sinB) = 0

Proof: Recall that sinA = a/2R, sinB = b/2R, and sinC = c/2R. By substitution, a(sinB - sinC) + b(sinC - sinA) + c(sinA - sinB) becomes

Rearranging and combining like terms:


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