# Prove that (a^2 - b^2 - c^2 + 2bc)/(b^2 + 2bc + c ^2 - a^2) =((s-b)(s-c))/(s(s-a)) If a + b + c = 2s?

Feb 21, 2015

$\frac{{a}^{2} - {b}^{2} - {c}^{2} + 2 b c}{{b}^{2} + 2 b c + {c}^{2} - {a}^{2}} = \left(\frac{{a}^{2} - \left({b}^{2} + {c}^{2} - 2 b c\right)}{\left({b}^{2} + 2 b c + {c}^{2}\right) - {a}^{2}}\right) =$

$= \left(\frac{{a}^{2} - {\left(b - c\right)}^{2}}{{\left(b + c\right)}^{2} - {a}^{2}}\right) = \left(1\right)$

At the numerator and at the denominator there is a difference of two squares, and remembering that:

${x}^{2} - {y}^{2} = \left(x - y\right) \left(x + y\right)$

$\left(1\right) = \frac{\left(a - \left(b - c\right)\right) \left(a + \left(b - c\right)\right)}{\left(\left(b + c\right) - a\right) \left(\left(b + c\right) + a\right)} = \frac{\left(a - b + c\right) \left(a + b - c\right)}{\left(b + c - a\right) \left(b + c + a\right)} = \left(2\right)$.

Now, since $s = \frac{1}{2} \left(a + b + c\right)$, than:

((s-b)(s-c))/(s(s-a))=((1/2(a+b+c)-b)(1/2(a+b+c)-c))/((1/2(a+b+c)(1/2(a+b+c)-a))=

$= \frac{\frac{a + b + c - 2 b}{2} \cdot \frac{a + b + c - 2 c}{2}}{\frac{1}{2} \left(a + b + c\right) \frac{a + b + c - 2 a}{2}} = \frac{\frac{1}{4} \left(a - b + c\right) \left(a + b - c\right)}{\frac{1}{4} \left(a + b + c\right) \left(b + c - a\right)} = \left(3\right)$

and simplifying for $\frac{1}{4}$ $\Rightarrow \left(2\right) = \left(3\right)$.