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Question
In triangle ABC, prove the following:
Solution
Let
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Then,
Consider the LHS of the equation
\[LHS = \frac{a^2 \sin\left( B - C \right)}{\sin A} + \frac{b^2 \sin\left( C - A \right)}{\sin B} + \frac{c^2 \sin\left( A - B \right)}{\sin C}\]
\[ = \frac{k^2 \sin^2 A\sin\left( B - C \right)}{\sin A} + \frac{k^2 \sin^2 B\sin\left( C - A \right)}{\sin B} + \frac{k^2 \sin^2 C\sin\left( A - B \right)}{\sin C} \]
\[ = k^2 \sin A\sin\left( B - C \right) + k^2 \sin B\sin\left( C - A \right) + k^2 \sin C\sin\left( A - B \right) \]
\[ = k^2 \left[ \sin A\left( \sin B\cos C - \sin C\cos B \right) + \sin B\left( \sin C\cos A - \sin A\cos C \right) + \sin C\left( \sin A\cos B - \sin B\cos A \right) \right] \]
\[ = k^2 \left( \sin A\sin B\cos C - \sin A\sin C\cos B + \sin B\sin C\cos A - \sin A\sin B\cos C + \sin A\sin C\cos B - \sin C\sin B\cos A \right)\]
\[ = 0 = RHS\]
\[\text{ Hence proved } .\]
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