Cho bốn điểm \(A,\,\,B,\,\,C,\,\,D\) trong mặt phẳng. Chứng minh rằng
\(\overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {BC} .\overrightarrow {AD} + \overrightarrow {CA} .\overrightarrow {BD} = 0.\)
Ta có: \(\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} \)
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\(\overrightarrow {CA} = \overrightarrow {BA} - \overrightarrow {BC} \)
\(\overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {CD} \)
Ta có:
\(\overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {BC} .\overrightarrow {AD} + \overrightarrow {CA} .\overrightarrow {BD} = \overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {BC} \left( {\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} } \right) + \left( {\overrightarrow {BA} - \overrightarrow {BC} } \right)\left( {\overrightarrow {BC} + \overrightarrow {CD} } \right)\)
\( = \overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {BC} .\overrightarrow {AB} + {\overrightarrow {BC} ^2} + \overrightarrow {BC} .\overrightarrow {CD} + \overrightarrow {BA} .\overrightarrow {BC} + \overrightarrow {BA} .\overrightarrow {CD} - {\overrightarrow {BC} ^2} - \overrightarrow {BC} .\overrightarrow {CD} \)
\( = \left( {\overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {BA} .\overrightarrow {CD} } \right) + \left( {\overrightarrow {BC} .\overrightarrow {AB} + \overrightarrow {BA} .\overrightarrow {BC} } \right) + \left( {{{\overrightarrow {BC} }^2} - {{\overrightarrow {BC} }^2}} \right) + \left( {\overrightarrow {BC} .\overrightarrow {CD} - \overrightarrow {BC} .\overrightarrow {CD} } \right) = 0\)
\( \Rightarrow \,\,\overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {BC} .\overrightarrow {AD} + \overrightarrow {CA} .\overrightarrow {BD} = 0\) (đpcm)