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a) Let $A$ and $B$ be 2 points in the plane. Show that if $M$ is the midpoint of the line segment $\overline {AB}$, then $\vec{OM} = \frac{1}{2} (\vec{OA}+\vec{OB})$ where $O$ is the origin.

I think I've got this one.

Let $A = (a_1, a_2) => \vec{OA} = (a_1, a_2)$

Let $B = (b_1, b_2) => \vec{OB} = (b_1, b_2)$

$M = ({\frac{a_1 + b_1}{2}},{\frac{a_2+b_2}{2}}) = \frac{1}{2}(a_1 + b_1, a_2 + b_2) = \frac{1}{2}[(a_1, a_2) + (b_1,b_2)] = \frac{1}{2}(\vec{OA} + \vec{OB})$

b) Use the formula for the midpoint which you proved in (a) to show that the midpoints of the 4 sides of any quadrilateral are the vertices of a parallelogram.

So if we have a quadrilateral $ABCD$ with $A=(a_1, a_2), B=(b_1,b_2), C=(c_1,c_2), D=(d_1,d_2).$

$M_{AB} = \frac{1}{2} (\vec{OA} + \vec{OB})$

$M_{BC} = \frac{1}{2} (\vec{OB} + \vec{OC})$

$M_{CD} = \frac{1}{2} (\vec{OC} + \vec{OD})$

$M_{AD} = \frac{1}{2} (\vec{OA} + \vec{OD})$

So now what is the most efficient way to take this further. Do I need to use the distance formula to prove that $M_{AB} M_{BC} = M_{AD} M_{CD}$ and $M_{AD} M_{AB} = M_{CD} M_{BC}$ as well as the fact that the product of gradients is -1 which proves that the lines are parallel

StephanCasey
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1 Answers1

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Now check that the vector from $M_{AB}$ to $M_{BC}$ is equal to the vector from $M_{AD}$ to $M_{CD}$, and that the vectors for the two other sides of the supposed parallelogram are equal. Equal vectors are parallel and have equal length.

Mark Bennet
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  • Oh thanks. That makes sense. So all I do is subtract the one from the other in both cases and it's done. It proves that they're parallel as well – StephanCasey Sep 27 '14 at 19:45
  • Let $\underline{x} = M_{BC} - M_{AB}$ and $\underline{y} = M_{AD} - M_{CD} => \underline{x} = \frac{1}{2}(\vec{OC} - \vec{OA}) and \underline{y} = \frac{1}{2}(\vec{OC} - \vec{OA}) => \underline{x} = \underline{y}.$ These sides are parallel and have equal length. Do you think I need to do the same thing for the other 2 sides or is this sufficient – StephanCasey Sep 27 '14 at 19:53
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    @StephanCasey In fact it is sufficient - if two opposite sides of a quadrilateral are parallel and have equal length, then the quadrilateral is a parallelogram. But you might want to prove that here (and it is easy) – Mark Bennet Sep 27 '14 at 20:01
  • Yeah. I will prove it in the test. Just in case. I'll also draw a diagram to illustrate – StephanCasey Sep 28 '14 at 09:40