# If a, b, c are real numbers such that a+b+c=2s, prove that a^2-b^2-c^2+2bc=4 (s-b) (s-c)

2
by sunayana1

Thanks a lot

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by sunayana1

Thanks a lot

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a^2 - b^2 - c^2 + 2bc = 4(s^2 - sc - sb + bc)

a^2 - b^2 - c^2 + 2bc = 4s^2 - 4sc - 4sb + 4bc

a^2 - b^2 - c^2 = (2s)^2 - 2c.2s - 2b.2s + 4bc - 2bc

since, a + b + c = 2s (given)

therefore,

a^2 - b^2 - c^2 = (a+b+c)^2 - 2c(a+b+c) - 2b(a+b+c) + 4bc - 2bc

a^2 - b^2 - c^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac - 2ac - 2bc - 2c^2 - 2ab - 2b^2 - 2bc + 2bc

- b^2 - c^2 = a^2 - a^2 + b^2 + c^2 - 2c^2 - 2b^2

- b^2 - c^2 = - b^2 - c^2 [PROVED]

a^2-b^2-c^2+2bc

=a^2-(b-c)^2

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

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

=(2s-2c)(2s-2b)

=4(s-c)(s-b)=RHS....(proved)