Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.

Question

My Solution

B (Conclusion): The midpoints of the sides of a space quadrilateral form a parallelogram.

A (Hypothesis): Let $A$, $B$, $C$, $D$ be four points such that they form a space quadrilateral.

B1: $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ where $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B}$ and $\dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ are congruent sides. The same can be said for the other two sides.

A1: $\mathbf{A} + \mathbf{B} = \mathbf{C} + \mathbf{D}$ by the definition of quadrilaterals.

$\implies \dfrac{1}{2} \left( \mathbf{A} + \mathbf{B} \right) = \dfrac{1}{2} \left( \mathbf{C} + \mathbf{D} \right)$

$\implies \dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$

$Q.E.D.$

I would greatly appreciate it if people could please review my proof for correctness.

I'm going to assume that a space quadrilateral is equivalent to a parallelogram No, in this context a "space quadrilateral" means an arbitrary quadrilateral which is not necessary planar (therefore "space", though that's not too common a usage). — dxiv, Jul 11 '17 at 19:46
I expect a space quadrilateral is a quadrilateral in an arbitrary-dimension vector space (i.e., a sequence of 4 distinct points). — John Hughes, Jul 11 '17 at 19:47
I have edited the OP with a diagram. @dxiv, if you're correct, then wouldn't a square also be a valid space quadrilateral? But the midpoints of the sides of a square aren't a parallelogram -- they're a square, aren't they? — The Pointer, Jul 11 '17 at 19:50
@ThePointer A square is a parallelogram. Rectangles, rhombi, squares are particular types of parallelograms. — dxiv, Jul 11 '17 at 19:53
@ThePointer If $A,B,C,D$ are the side vectors (rather than position vectors of the vertices), then $A+B=C+D$ holds true for any quadrilateral, so you don't need the parallelogram assumption. — dxiv, Jul 11 '17 at 20:07
@ThePointer B1: ... where ... are congruent sides You still need to justify that $(A+B)/2$ is in fact a side of the parallelogram formed by the midpoints (even if it may be obvious). — dxiv, Jul 11 '17 at 21:48

score 1 · Answer 1 · answered Jul 11 '17 at 19:51

Hint: If your four points are $a, b, c, d$, then the midpoints, in order around the quad, are $$ p = \frac{1}{2}(a+b), q = \frac{1}{2}(b+c), r = \frac{1}{2}(c+d), s = \frac{1}{2}(d+a). $$

For $pqrs$ to be a parallelogram, you need the edge from $p$ to $q$ to have the same direction vector as the edge from $s$ to $r$; you need a similar thing to hold for the edges from $q$ to $r$ and $p$ to $s$.

What's the direction vector of the edge from $p$ to $q$? Can you express it in terms of $a, b, c, d$?

score 1 · Answer 2 · answered Jul 11 '17 at 19:59

1

Hint:

since $\vec A+\vec B=\vec C+ \vec D$ we have:

$$ \frac{1}{2}(\vec A+\vec B)=\frac{1}{2}(\vec C+ \vec D)\iff \frac{1}{2}\vec A +\frac{1}{2}\vec B=\frac{1}{2}\vec C+\frac{1}{2}\vec D $$

answered Jul 11 '17 at 19:59

Emilio Novati

62,675

But isn't this what I did? – The Pointer Jul 11 '17 at 19:59
1

Yes it is essentally the same, and it is true for any for vectors ( in any vector space) such that $\vec A+\vec B=\vec C+ \vec D$ . You only have to better formulate the hypotesis and the conclusion, And the figure is misleading because the statement is true also for non coplanar vectors. – Emilio Novati Jul 11 '17 at 20:06
Ok, thanks. What do you mean by "better formulate the hypothesis and conclusion"? – The Pointer Jul 11 '17 at 20:08
1

Hypotesis: Let $A,B,C,D$ be four points such that form a quadrilateral (not a parallelogram) in an affine space. – Emilio Novati Jul 11 '17 at 20:13
Ok. thanks. So everything else is ok? – The Pointer Jul 11 '17 at 20:15
Properly, you $B$ is not the ''Conclusion'' but the statement of the ''theorem'' ( but I'm not native English speaker and maybe that I dont known the correct words) – Emilio Novati Jul 11 '17 at 20:21
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@ThePointer You wrote these are the aforementioned midpoints in the question, but that's wrong. $(A+B)/2$ is not the midpoint of any side of the quadrilateral in the posted figure, which seems to imply that $A,B,C,D$ are the vectors corresponding to the sides of the quadrilateral. In that case $(A+B)/2$ is simply half of the diagonal from the tail of $A$ to the head of $B$. You should really clarify in the question what exactly $A,B,C,D$ are supposed to be. – dxiv Jul 11 '17 at 20:21
@dxiv You're right -- I seem to have made an error. – The Pointer Jul 11 '17 at 20:31
@dxiv Ok, I think I fixed it. – The Pointer Jul 11 '17 at 21:27

score 0 · Answer 3 · answered Jul 11 '17 at 19:42

In general, the midpoints of any convex quadrilateral form a parallelogram, and you can prove that quite easily by drawing diagonals of the initial quadrilateral, but I'm not exactly sure what a space parallelogram is either, nor do I know how to prove this using vectors or check your proof as I have close to none understanding of them.

Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.

3 Answers3