Prove that $\frac{a^{2} - b^{2} - c^{2} + 2 b c}{b^{2} + 2 b c + c^{2} - a^{2}} = \frac{(s - b) (s - c)}{s (s - a)}$ $(a^2 - b^2 - c^2 + 2bc)/(b^2 + 2bc + c ^2 - a^2) =((s-b)(s-c))/(s(s-a))$ If $a + b + c = 2 s$ $a + b + c = 2s$ ?

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Prove that $\frac{a^{2} - b^{2} - c^{2} + 2 b c}{b^{2} + 2 b c + c^{2} - a^{2}} = \frac{(s - b) (s - c)}{s (s - a)}$ $(a^2 - b^2 - c^2 + 2bc)/(b^2 + 2bc + c ^2 - a^2) =((s-b)(s-c))/(s(s-a))$ If $a + b + c = 2 s$ $a + b + c = 2s$ ?

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Answer 1 · 2015-02-21T14:56:41

Follow my passages:

$\frac{a^{2} - b^{2} - c^{2} + 2 b c}{b^{2} + 2 b c + c^{2} - a^{2}} = (\frac{a^{2} - (b^{2} + c^{2} - 2 b c)}{(b^{2} + 2 b c + c^{2}) - a^{2}}) =$ $(a^2-b^2-c^2+2bc)/(b^2+2bc+c^2-a^2)=((a^2-(b^2+c^2-2bc))/((b^2+2bc+c^2)-a^2))=$

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

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

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

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

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

$\frac{(s - b) (s - c)}{s (s - a)} = \frac{(\frac{1}{2} (a + b + c) - b) (\frac{1}{2} (a + b + c) - c)}{(\frac{1}{2} (a + b + c) (\frac{1}{2} (a + b + c) - a)) =}$ $((s-b)(s-c))/(s(s-a))=((1/2(a+b+c)-b)(1/2(a+b+c)-c))/((1/2(a+b+c)(1/2(a+b+c)-a))=$

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

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

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Prove that $\frac{a^{2} - b^{2} - c^{2} + 2 b c}{b^{2} + 2 b c + c^{2} - a^{2}} = \frac{(s - b) (s - c)}{s (s - a)}$ $(a^2 - b^2 - c^2 + 2bc)/(b^2 + 2bc + c ^2 - a^2) =((s-b)(s-c))/(s(s-a))$ If $a + b + c = 2 s$ $a + b + c = 2s$ ?

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Prove that $\frac{a^{2} - b^{2} - c^{2} + 2 b c}{b^{2} + 2 b c + c^{2} - a^{2}} = \frac{(s - b) (s - c)}{s (s - a)}$ $(a^2 - b^2 - c^2 + 2bc)/(b^2 + 2bc + c ^2 - a^2) =((s-b)(s-c))/(s(s-a))$ If $a + b + c = 2 s$ $a + b + c = 2s$ ?