Đề bài

Cho tam giác \(ABC.\) Chứng minh rằng: a)      \(\cot A + \cot B + \cot C = \frac{{{a^2} + {b^2} + {c^2}}}{{4S}}.\) b)     \(m_a^2 + m_b^2 + m_c^2 = \frac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right).\)

Lời giải chi tiết

a) \(\cot A + \cot B + \cot C = \frac{{{a^2} + {b^2} + {c^2}}}{{4S}}.\)\(\begin{array}{l}VT = \frac{{\cos A}}{{\sin A}} + \frac{{\cos B}}{{\sin B}} + \frac{{\cos C}}{{\sin C}} = \frac{{\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{{\frac{{2S}}{{bc}}}} + \frac{{\frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}}}{{\frac{{2S}}{{ac}}}} + \frac{{\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}}}{{\frac{{2S}}{{ab}}}}\\ = \frac{{{b^2} + {c^2} - {a^2}}}{{4S}} + \frac{{{a^2} + {c^2} - {b^2}}}{{4S}} + \frac{{{a^2} + {b^2} - {c^2}}}{{4S}}\\ = \frac{{{a^2} + {b^2} + {c^2}}}{{4S}} = VP\,\,\left( {dpcm} \right)\end{array}\)b) \(m_a^2 + m_b^2 + m_c^2 = \frac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right).\)

\(\begin{array}{l}VT = \left( {\frac{{{b^2} + {c^2}}}{2} - \frac{{{a^2}}}{4}} \right) + \left( {\frac{{{a^2} + {c^2}}}{2} - \frac{{{b^2}}}{4}} \right) + \left( {\frac{{{a^2} + {b^2}}}{2} - \frac{{{c^2}}}{4}} \right)\\ = \frac{{2\left( {{a^2} + {b^2} + {c^2}} \right)}}{2} - \frac{{{a^2} + {b^2} + {c^2}}}{4}\\ = \frac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right) = VP\,\,\left( {dpcm} \right).\end{array}\)