# How to draw $(A,B)\sim (C,D) \implies (A,C)\sim (B,D)$ when $A,B,C,D$ are collinears?

$(A,B)$ and $(C,D)$ are parallel vectors, in the book I'm reading, it illustrates one case for this proposition: $(A,B)\sim (C,D) \implies (A,C)\sim (B,D)$ with the following figure:

And then there's an exercise asking me to draw that proposition in the case that $A,B,C,D$ are collinear. I'm not sure about the answer but I guess that to achieve that, I should think of null vectors. The only possibility I see to achieve collinearity between all them is to think of $A=B=C=D$. Is this the answer?

Edit: I guess this is a viable answer:

The vector $(A,B)$ approach to $(C,D)$ until one is over the other.

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What are you talking about? – copper.hat Dec 13 '12 at 17:43
Is it better now? – Voyska Dec 13 '12 at 18:44
Excuse the holes in my education, but what does ~ mean? – copper.hat Dec 13 '12 at 18:46
@copper.hat $\sim$ means similar. About the holes on your education, there are only two reasons to be here: 1 - Filling these holes with knowing; 2 - Help others to do so. – Voyska Dec 13 '12 at 18:56
What other rules are satisfied? Is $(A,B)+(B,D) \sim (A,D)$? – copper.hat Dec 13 '12 at 18:59

Ok, so correct me if I'm wrong, but I assume by collinear that just means points A, B, C, and D all fall on the same line.

So, assume they all fall on a straight line:

Suppose vector AB is magnitude +3, similar to CD. Then suppose on the line A,B,C, and D fall on there's X magnitude between BC. It follows that AC is 3+X and BD is 3+X, hence AC~BD?

From Wikipedia:

In geometry, collinearity is a property of a set of points, specifically, the property of lying on a single line.

In other words, a bunch of points that fall on the same line, e.g. are aligned:

Now if A, B, C, and D are collinear (as above), and the distance (magnitude) of $\overrightarrow {AB}$ is the same (~) as $\overrightarrow {CD}$ then you can see, visually that it follows that $\overrightarrow {AC}$ ~ $\overrightarrow {BD}$

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Trying to understand. Did you see the edit in my post? – Voyska Dec 15 '12 at 14:02
But shouldn't you draw a case in which $A,B,C,D$ are collinear? Is this the case? – Voyska Dec 15 '12 at 16:11