\(\left\{ {\begin{array}{*{20}{l}}{\overrightarrow {IJ} = \overrightarrow {IA} + \overrightarrow {AB} + \overrightarrow {BJ} }\\{\overrightarrow {IJ} = \overrightarrow {ID} + \overrightarrow {DC} + \overrightarrow {CJ} }\end{array}} \right.\)

Cộng vế với vế:

\(2\overrightarrow {IJ} = \left( {\overrightarrow {IA} + \overrightarrow {ID} } \right) + \left( {\overrightarrow {BJ} + \overrightarrow {CJ} } \right) + \overrightarrow {AB} + \overrightarrow {DC} = \overrightarrow {AB} + \overrightarrow {DC} \)

\( \Rightarrow \overrightarrow {IJ} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {DC} \) 

b) Đặt \(\frac = \frac = k\)

\(\left\{ {\begin{array}{*{20}{c}}{\overrightarrow {IP} = \overrightarrow {IA} + \overrightarrow {AM} + \overrightarrow {MP} }\\{\overrightarrow {IP} = \overrightarrow {ID} + \overrightarrow {DN} + \overrightarrow {NP} }\end{array}} \right.\)

\( \Rightarrow 2\overrightarrow {IP} = \left( {\overrightarrow {IA} + \overrightarrow {ID} } \right) + \left( {\overrightarrow {MP} + \overrightarrow {NP} } \right) + \overrightarrow {AM} + \overrightarrow {DN} = \overrightarrow {AM} + \overrightarrow {DN} \)

\( \Rightarrow 2\overrightarrow {IP} = k \cdot \overrightarrow {AB} + k \cdot \overrightarrow {DC} \)

\( \Rightarrow \overrightarrow {IP} = \frac{k}{2}\left( {\overrightarrow {AB} + \overrightarrow {DC} } \right) = \frac{k}{2} \cdot \overrightarrow {IJ} \)

⇒ P; I; J thẳng hàng hay P thuộc IJ.