a) Ta có:
\(\overrightarrow {BC} + \overrightarrow {NM} = \overrightarrow {BM} + \overrightarrow {MC} + \overrightarrow {NC} + \overrightarrow {CM} = \overrightarrow {BM} + \overrightarrow {NC} \,\,\,\left( {dpcm} \right).\)
b) Biểu diễn \(\overrightarrow {AI} \) theo \(\overrightarrow {AB} ,\,\,\overrightarrow {AC} .\)
Ta có:
\(\overrightarrow {NC} = 2\overrightarrow {NA} \Rightarrow \left\{ \begin{array}{l}\overrightarrow {NA} = \frac{1}{2}\overrightarrow {AC} \\\overrightarrow {NC} = \frac{3}{2}\overrightarrow {AC} \end{array} \right..\)
\(\begin{array}{l}\overrightarrow {AI} = \overrightarrow {AN} + \overrightarrow {NI} = - \frac{1}{2}\overrightarrow {AC} + \frac{1}{2}\overrightarrow {NM} \\ = - \frac{1}{2}\overrightarrow {AC} + \frac{1}{2}\left( {\overrightarrow {BM} + \overrightarrow {NC} - \overrightarrow {BC} } \right)\\ = - \frac{1}{2}\overrightarrow {AC} - \frac{1}{2}.\frac{1}{2}\overrightarrow {AB} + \frac{1}{2}.\frac{3}{2}\overrightarrow {AC} - \frac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {AC} } \right)\\ = - \frac{1}{2}\overrightarrow {AC} - \frac{1}{4}\overrightarrow {AB} + \frac{3}{4}\overrightarrow {AC} + \frac{1}{2}\overrightarrow {AB} - \frac{1}{2}\overrightarrow {AC} \\ = \frac{1}{4}\overrightarrow {AB} - \frac{1}{4}\overrightarrow {AC} .\end{array}\)
