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Prove that the quadrilateral with vectors $A, B, C, D$ is a parallelogram if and only if

$$A-B+C-D = 0$$


It is my understanding that it is a parallelogram if the opposing sides of the quadrilateral are parallel and have the same length.

$$A-B+C-D = 0$$

Is basically

$$\vec{BA}+\vec{DC} = 0$$

$$\vec{BA} = \vec{CD}$$

We now know that there are two sides that are equal... wait, no, not necessarily I think:

$\vec{CD}$ could be the quadrilateral's diagonal, if points $C$ and $D$ were on opposing corners, right?

How do I proceed?

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You began your problem statement with the phrase "Prove that the quadrilateral $ABCD$ is...". In order for that phrase to make sense, $ABCD$ cannot refer to more than one quadrilateral. If there was more than one, we'd have to say "Prove that a quadrilateral $ABCD$ is..." instead.

It all depends on whether or not the representation $ABCD$ indicates the order (which is usually the case) in which the vertices are connected, in addition to the vertices themselves. Knowing just the set of vertices, the quadrilateral may either be simple or self-intersecting, depending on how you connect the vertices. So since the problem statement seems to be presuming that $ABCD$ unambiguously refers to a single quadrilateral, we can assume that the vertices $A,B,C,D$ are connected in that order.

This makes your question a non-issue. You ask what to do in the case that $D$ is opposite $C$, and the answer is that $D$ never is opposite $C$. The point $A$ is.

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You have proved that $\vec{BA}$ is parallel, and of the same length as, $\vec{CD}$. A similar argument will prove that $\vec{AD}$ is parallel, and of the same length as, $\vec{BC}$. Now draw the quadrilateral and make use of all this information.

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