91Ó°ÊÓ

The Law of Cosines states that, in any triangle with sides a, b, c and their respective opposite angles A, B, C: \[a^2=b^2+c^2-2bc\cos A\] \[b^2=a^2+c^2-2ac\cos B\] \[c^2=a^2+b^2-2ab\cos C\] We will use these laws to express the cosine of angles A, B, and C. Solve for \(\cos A\), \(\cos B\), and \(\cos C\), respectively: \[\cos A = \frac{b^2+c^2-a^2}{2bc}\] \[\cos B = \frac{a^2+c^2-b^2}{2ac}\] \[\cos C = \frac{a^2+b^2-c^2}{2ab}\]

Now, we'll replace \(\cos A\), \(\cos B\), and \(\cos C\) in the given equation with the expressions we found in Step 1: \((b+c)\left(\frac{b^2+c^2-a^2}{2bc}\right) + (c+a)\left(\frac{a^2+c^2-b^2}{2ac}\right) + (a+b)\left(\frac{a^2+b^2-c^2}{2ab}\right) = a+b+c\)

To simplify the equation, the denominators are all a product of the sides (e.g., 2bc, 2ac, 2ab). Therefore, the denominators can cancel out with the corresponding "side expressions" in the numerators: \[\frac{(b+c)(b^2+c^2-a^2)}{2bc} + \frac{(c+a)(a^2+c^2-b^2)}{2ac} + \frac{(a+b)(a^2+b^2-c^2)}{2ab} = a+b+c\] \[\frac{b^3+c^3-a^2b-a^2c}{2} + \frac{a^3+c^3-b^2a-b^2c}{2} + \frac{a^3+b^3-c^2a-c^2b}{2} = a+b+c\] Now combine like terms: \[\frac{a^3+b^3+c^3-a^2b-a^2c-b^2a-b^2c-c^2a-c^2b}{2} = a+b+c\] \[\frac{a^3-a^2b-a^2c+b^3-b^2a-b^2c+c^3-c^2a-c^2b}{2} = a+b+c\] Factor out a common factor from each term: \[a\left(\frac{a-b-c}{2}\right) + b\left(\frac{b-a-c}{2}\right) + c\left(\frac{c-a-b}{2}\right) = a+b+c\] We can observe that all three factors in parentheses are equal to -1: \[a(-1) + b(-1) + c(-1) = a+b+c\] Simplify this equation: \[-a-b-c = a+b+c\] Finally, add a, b, and c to both sides of the equation: \[0 = 2(a+b+c)\] Since adding zero to a quantity leaves it unchanged, our final equation is: \[a+b+c = a+b+c\] The given equation is proven true.