# Prove $(b-c)\sin A+(c-a)\sin B+(a-b)\sin C=0$ [closed]

Prove the following equation, when you consider it as $BC=a$, $CA=b$, and $AB=c$ in a triangle　$ABC$.

$(b-c)\sin A+(c-a)\sin B+(a-b)\sin C=0$

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## closed as off-topic by Matt Pressland, Daniel Rust, Lord_Farin, Davide Giraudo, Tom OldfieldOct 9 at 11:31

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Using this or this, $$\frac b{\sin B}=\frac c{\sin C}=\frac a{\sin A}=2R$$
So, $$(b-c)\sin A=(2R\sin B-2R\sin C)\sin A=2R(\sin A\sin B-\sin C\sin A)$$
or $$(b-c)\sin A=(b-c)\frac a{2R}=\frac{ab-ca}{2R}$$