Cho tứ giác ABCD.

Chứng minh rằng nếu \(\left| {\overrightarrow {A{\rm{D}}} + \overrightarrow {BC} } \right| = \left| {\overrightarrow {AB} + \overrightarrow {DC} } \right|\) thì AC ⊥ BD.

Ta có:

• \(\overrightarrow {A{\rm{D}}} + \overrightarrow {BC} = \left( {\overrightarrow {AC} + \overrightarrow {CD} } \right) + \left( {\overrightarrow {BD} + \overrightarrow {DC} } \right)\)

                  \( = \overrightarrow {AC} + \overrightarrow {BD} + \left( {\overrightarrow {CD} + \overrightarrow {DC} } \right) = \overrightarrow {AC} + \overrightarrow {BD} \)

• \(\overrightarrow {AB} + \overrightarrow {DC} = \left( {\overrightarrow {AC} + \overrightarrow {CB} } \right) + \left( {\overrightarrow {DB} + \overrightarrow {BC} } \right)\)

                     \( = \overrightarrow {AC} + \overrightarrow {DB} + \left( {\overrightarrow {CB} + \overrightarrow {BC} } \right) = \overrightarrow {AC} + \overrightarrow {DB} \)

Theo bài: \(\left| {\overrightarrow {A{\rm{D}}} + \overrightarrow {BC} } \right| = \left| {\overrightarrow {AB} + \overrightarrow {DC} } \right|\)

\[ \Leftrightarrow \left| {\overrightarrow {AC} + \overrightarrow {BD} } \right| = \left| {\overrightarrow {AC} + \overrightarrow {DB} } \right|\]

\[ \Leftrightarrow {\left| {\overrightarrow {AC} + \overrightarrow {BD} } \right|^2} = {\left| {\overrightarrow {AC} + \overrightarrow {DB} } \right|^2}\]

\[ \Leftrightarrow {\overrightarrow {AC} ^2} + 2\overrightarrow {AC} .\overrightarrow {BD} + {\overrightarrow {BD} ^2} = {\overrightarrow {AC} ^2} + 2\overrightarrow {AC} .\overrightarrow {DB} + {\overrightarrow {DB} ^2}\]

\[ \Leftrightarrow 2\overrightarrow {AC} .\overrightarrow {BD} + {\overrightarrow {BD} ^2} = - 2\overrightarrow {AC} .\overrightarrow {BD} + {\overrightarrow {BD} ^2}\]

\[ \Leftrightarrow 4\overrightarrow {AC} .\overrightarrow {BD} = 0\]

\[ \Leftrightarrow \overrightarrow {AC} .\overrightarrow {BD} = 0\]

\[ \Leftrightarrow \overrightarrow {AC} \bot \overrightarrow {BD} \]

Suy ra AC ⊥ BD.