Cho hình bình hành ABCD. Lấy các điểm M, N, P thỏa mãn \[\overrightarrow {AM} = \frac{1}{2}\overrightarrow {AB} ,\,\,\overrightarrow {AN} = \frac{1}{5}\overrightarrow {AC} ,\,\,\overrightarrow {AP} = \frac{1}{3}\overrightarrow {AD} \]. Đặt \[\overrightarrow {AB} = \overrightarrow a ,\,\,\overrightarrow {AD} = \overrightarrow b \]. Chứng minh ba điểm M, N, P thẳng hàng.

Ta có:

• \[\overrightarrow {AN} = \frac{1}{5}\overrightarrow {AC} = \frac{1}{5}\left( {\overrightarrow {AB} + \overrightarrow {AD} } \right)\]

\[ = \frac{1}{5}\overrightarrow {AB} + \frac{1}{5}\overrightarrow {AD} = \frac{1}{5}\vec a + \frac{1}{5}\vec b\]

• \[\overrightarrow {MN} = \overrightarrow {AN} - \overrightarrow {AM} = \frac{1}{5}\overrightarrow {AC} - \frac{1}{2}\overrightarrow {AB} \]

\[ = \frac{1}{5}\left( {\overrightarrow {AB} + \overrightarrow {AD} } \right) - \frac{1}{2}\overrightarrow {AB} \]

\[ = - \frac{3}{{10}}\overrightarrow {AB} + \frac{1}{5}\overrightarrow {AD} = - \frac{3}{{10}}\vec a + \frac{1}{5}\vec b\]

• \[\overrightarrow {NP} = \overrightarrow {AP} - \overrightarrow {AN} = \frac{1}{3}\overrightarrow {AD} - \frac{1}{5}\overrightarrow {AC} \]

\[ = \frac{1}{3}\overrightarrow {AD} - \frac{1}{5}\left( {\overrightarrow {AB} + \overrightarrow {AD} } \right)\]

\[ = - \frac{1}{5}\overrightarrow {AB} + \frac{2}{{15}}\overrightarrow {AD} = - \frac{1}{5}\vec a + \frac{2}{{15}}\vec b\]

Ta có: \[ - \frac{3}{{10}}\vec a + \frac{1}{5}\vec b = \frac{3}{2}\left( { - \frac{1}{5}\vec a + \frac{2}{{15}}\vec b} \right)\] hay \[\overrightarrow {MN} = \frac{3}{2}\overrightarrow {NP} \].

Vậy M, N, P thẳng hàng.