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))=a2−b2−c2+2bcb2+2bc+c2−a2=(a2−(b2+c2−2bc)(b2+2bc+c2)−a2)=
=((a^2-(b-c)^2)/((b+c)^2-a^2))=(1)=(a2−(b−c)2(b+c)2−a2)=(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)x2−y2=(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)(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)s=12(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))=(s−b)(s−c)s(s−a)=(12(a+b+c)−b)(12(a+b+c)−c)(12(a+b+c)(12(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)=a+b+c−2b2⋅a+b+c−2c212(a+b+c)a+b+c−2a2=14(a−b+c)(a+b−c)14(a+b+c)(b+c−a)=(3)
and simplifying for 1/414 rArr(2)=(3)⇒(2)=(3).
