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#?

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Answer 1 · 2015-02-21T14:56:41

Follow my passages:

#(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))=#

#=((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)#

#(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=1/2(a+b+c)#, 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))=#

#=((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 #1/4# #rArr(2)=(3)#.