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Question

For any triangle ABC prove that 2(bccosA+cacosB+abcosC)=(a2+b2+c2).

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Solution

Now for any ΔABC,
2(bccosA+cacosB+abcosC)
=2(bccosA+cacosB+abcosC)
=(b2+c2a2)+(a2+c2a2)+(a2+b2c2) [ Since cosA=b2+c2a22bc and similarly for cosB,cosC]
=(a2+b2+c2)

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