a)
\(\begin{array}{l}\overrightarrow {AB} + \overrightarrow {CD} = \overrightarrow {AD} - \overrightarrow {BC} \\ \Leftrightarrow \overrightarrow {AB} - \overrightarrow {AD} = - \overrightarrow {CD} - \overrightarrow {BC} \\ \Leftrightarrow \overrightarrow {DB} = \overrightarrow {CB} - \overrightarrow {CD} \\ \Leftrightarrow \overrightarrow {DB} = \overrightarrow {DB} \left( {dung} \right) \Rightarrow dpcm\end{array}\)b) Ta có:
\(\begin{array}{l}\overrightarrow {AG} = \dfrac{1}{3}\left( {\overrightarrow {AA} + \overrightarrow {AB} + \overrightarrow {AC} } \right) = \dfrac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\\\overrightarrow {MN} = \overrightarrow {AN} - \overrightarrow {AM} = \dfrac{2}{5}\overrightarrow {AC} - 2\overrightarrow {AB} \\\overrightarrow {MG} = \overrightarrow {AG} - \overrightarrow {AM} = \dfrac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) - 2\overrightarrow {AB} \\ = - \dfrac{5}{3}\overrightarrow {AB} + \dfrac{1}{3}\overrightarrow {AC} = \dfrac{5}{6}\left( {\dfrac{2}{5}\overrightarrow {AC} - 2\overrightarrow {AB} } \right) = \dfrac{5}{6}\overrightarrow {MN} \end{array}\)
\( \Rightarrow M,N,G\) thẳng hàng.
