Đặt \(\frac{{AM}}{{AD}} = \frac{{CN}}{{CB}}\)= k
Suy ra: \[\left\{ \begin{array}{l}\overrightarrow {AM} = \,k.\,\overrightarrow {AD} \\\overrightarrow {CN} = \,k.\,\overrightarrow {CB} \end{array} \right.\] với k là hằng số
\[\overrightarrow {EI} = \,\overrightarrow {EC} + \overrightarrow {CN} + \overrightarrow {NF\,} = \,\frac{1}{2}\overrightarrow {AC} + \overrightarrow {CN} + \frac{1}{2}\overrightarrow {NM} \]
\[\overrightarrow {EI} = \,\frac{1}{2}\left( {\overrightarrow {AD} + \overrightarrow {DC} } \right) + \,\overrightarrow {CN} + \frac{1}{2}\left( {\overrightarrow {NC} + \overrightarrow {CD} + \overrightarrow {DM} } \right)\,\]
\[ = \,\frac{1}{2}\overrightarrow {AD} + \frac{1}{2}\overrightarrow {DC} + \,\overrightarrow {CN} + \frac{1}{2}\overrightarrow {NC} + \frac{1}{2}\overrightarrow {CD} + \overrightarrow {DM} \]
= \[\frac{1}{2}\overrightarrow {AD} + \frac{1}{2}\overrightarrow {CN} + \frac{1}{2}\overrightarrow {DM} \, = \,\frac{1}{2}\overrightarrow {AM} + \frac{1}{2}\overrightarrow {CN} \, = \,\,\,\frac{k}{2}\left( {\overrightarrow {AD} + \overrightarrow {CB} } \right)\]
\[\overrightarrow {EF} = \,\overrightarrow {EC} + \overrightarrow {CB} + \overrightarrow {BF\,} = \,\frac{1}{2}\overrightarrow {AC} + \overrightarrow {CB} + \frac{1}{2}\overrightarrow {BD} \]
= \[\frac{1}{2}\overrightarrow {AD} + \frac{1}{2}\overrightarrow {DC} + \,\overrightarrow {CB} \, + \,\frac{1}{2}\overrightarrow {BC} + \frac{1}{2}\overrightarrow {CD} = \,\frac{1}{2}\overrightarrow {AD} + \frac{1}{2}\overrightarrow {CB} \, = \,\,\,\frac{1}{2}\left( {\overrightarrow {AD} + \overrightarrow {CB} } \right)\]
Suy ra: \(\overrightarrow {EF} \, = \,k.\,\overrightarrow {EI} \)
Vậy E, F, I thẳng hàng hay I luôn chuyển động trên đoạn EF.