Cho tam giác ABC với BC = a,AC = b,AB = c, p là nửa chu vi của tam giác. Chứng minh rằng abc cos A + cos B + cos C = {a^2} (p - a) + {b^2} {p - b} + {c^2} {p - c}

Ta có

\(abc\left( {\cos A + \cos B + \cos C} \right) = abc\left( {\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} + \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right)\)

\(\begin{array}{l} = a.\frac{{{b^2} + {c^2} - {a^2}}}{2} + b.\frac{{{a^2} + {c^2} - {b^2}}}{2} + c.\frac{{{a^2} + {b^2} - {c^2}}}{2}\\ = \frac{{{a^2}}}{2}\left( {b + c - a} \right) + \frac{{{b^2}}}{2}\left( {c + a - b} \right) + \frac{{{c^2}}}{2}\left( {a + b - c} \right)\end{array}\)\( = {a^2}\left( {p - a} \right) + {b^2}\left( {p - b} \right) + {c^2}\left( {p - c} \right)\).