Hướng dẫn trả lờiTheo quy tắc ba điểm, ta có\(\eqalign{
& \overrightarrow {AD} + \overrightarrow {BE} + \overrightarrow {CF} = \left( {\overrightarrow {AE} + \overrightarrow {ED} } \right) + \left( {\overrightarrow {BF} + \overrightarrow {FE} } \right) + \left( {\overrightarrow {CD} + \overrightarrow {DF} } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow {AE} + \overrightarrow {BF} + \overrightarrow {CD} + \left( {\overrightarrow {FE} + \overrightarrow {ED} + \overrightarrow {DF} } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow {AE} + \overrightarrow {BF} + \overrightarrow {CD} + \left( {\overrightarrow {FD} + \overrightarrow {DF} } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow {AE} + \overrightarrow {BF} + \overrightarrow {CD} \cr} \)Tương tự, ta cũng có\(\eqalign{
& \overrightarrow {AD} + \overrightarrow {BE} + \overrightarrow {CF} = \left( {\overrightarrow {AF} + \overrightarrow {FD} } \right) + \left( {\overrightarrow {BD} + \overrightarrow {DE} } \right) + \left( {\overrightarrow {CE} + \overrightarrow {EF} } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow {AF} + \overrightarrow {BD} + \overrightarrow {CE} + \left( {\overrightarrow {FD} + \overrightarrow {DE} + \overrightarrow {EF} } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow {AF} + \overrightarrow {BD} + \overrightarrow {CE} + \left( {\overrightarrow {FE} + \overrightarrow {EF} } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow {AF} + \overrightarrow {BD} + \overrightarrow {CE} \cr} \)Vậy ta có \(\overrightarrow {AD} + \overrightarrow {BE} + \overrightarrow {CF} = \overrightarrow {AE} + \overrightarrow {BF} + \overrightarrow {CD} = \overrightarrow {AF} + \overrightarrow {BD} + \overrightarrow {CE} \)