what kind of quadrilateral is ABCD?

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.

• 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. – GEdgar Sep 15 '11 at 0:58
• 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. – Henning Makholm Sep 15 '11 at 1:10
• 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). – Henning Makholm Sep 15 '11 at 1:29
• @Henning, that's great, thanks. – Charles Bao Sep 15 '11 at 1:36
• @Henning: Maybe you want to flesh that out into a full answer? – davidlowryduda Aug 21 '12 at 22:06

Some informal degrees-of-freedom analysis:

• An arbitrary quadrilateral on a plane is described by $8$ parameters: coordinates of each vertex (to simplify matter, I don't take quotient by isometries).
• A rectangle on a plane is described by $5$ parameters: endpoints of one side and (signed) length of the other side.

We should not expect two equations to restrict the $8$-dimensional space of quadrilaterals down to $5$-dimensional space of rectangles. Three equations (also given in the post) are enough.

The above is not a rigorous proof because two equations $f=0=g$ can be made into one $f^2+g^2=0$, etc. One needs some transversality consideration to make it work. But it's easier to just quote a geometric argument given by Henning Makholm in the comments.

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 $CD$.

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 $A(1,1)$, $B(1,2)$, $C(0,0)$, $D(0,2)$.

The first of Henning's examples is below (the second isn't even convex)