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ABCD is a quadrilateral, given $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}$, then what kind of quadrilateral is ABCD? I guess it's a rectangle, but how to prove it?

If the situation becomes $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}=\overrightarrow{DA}\cdot\overrightarrow{AB}$, i can easily prove ABCD is a rectangle.

So, the question is given $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}$, can we get $\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{CD}=\overrightarrow{CD}\cdot\overrightarrow{DA}=\overrightarrow{DA}\cdot\overrightarrow{AB}$?

thanks.

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    Why do you think these have a special name? And (for your definition of quadrilateral) it is unusual to allow the edges to cross each other, I think.2011-09-15
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    If you place B, C, and D arbitrarily, then each of the two equations between dot products defines a line that A must lie on. Putting A at the intersection between these two lines gives you a quadrilateral that satisfies the condition. So you cannot conclude anything about the angle at C (i.e., it _doesn't_ have to be a rectangle) -- nor anything about the relative lengths of BC versus CD.2011-09-15
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    A concrete example would be A(2,5), B(-1,1), C(0,0), D(1,0). Doesn't look like anything that has a nice name. Neither does A(1,1), B(1,2), C(0,0), D(0,2).2011-09-15
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    @Henning, that's great, thanks.2011-09-15
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    @Henning: Maybe you want to flesh that out into a full answer?2012-08-21

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