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Homework Help: Vector algebra (proving you have a parallelogram by using vectors)

  1. Nov 6, 2016 #1
    1. The problem statement, all variables and given/known data
    23. In a ABCD quadrilateral let P,Q,R,S be midpoints of sides AB,BC,CD and DA. Let X be the intersection of BR and DQ, and let Y be the intersection of BS and DP. If ##\vec{BX}=\vec{YD} ## show that ABCD is a parallelogram .

    Slika nove bitne slike.jpg
    2. Relevant equations
    ## (\vec{a}\cdot\vec{b})=0## then a and b are perpendicular
    I can't think of any other useful ones
    3. The attempt at a solution

    So when I first saw this problem I didn't really have an idea how to solve it so I just tried figuring out what are the rations that X divides BR into. I calculated that the ratio is 2/3. I did the same for Y and DP and also got the ratios 2/3. Which I though was good since I figured out that ##\vec{BX}=\vec{YD} ## really are the same. However I still didn't have any idea how to continue so then I decided to figure out ratios of AY,AX,XY and found out that AY/(AC)=XY/(AC)=XC/(AC)=1/3 however I don't see how any of this would help me show that ABCD is a parallelogram. I'm kinda lost here.
    I though about trying to show that ## (\vec{a}\cdot\vec{b})=/=0## since that would mean that a and b are not perpendicular and that we have a parallelogram however I was not able to find a way to show that.
    Any help is greatly appreciated.
  2. jcsd
  3. Nov 6, 2016 #2

    Simon Bridge

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    I'd have started with a diagram of a random quadralateral ... rather than a rectangle.
    You need to keep in mind the definition of a parallelogram ... play about with the diagrams, look at how the geometry differs for different parralellograms.
  4. Nov 10, 2016 #3


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    By definition, BR and DQ are medians in the green triangle BCD, X is the center of gravity and it divides the medians in the ratio 1/2. It is the same for Y in triangle ABD. The thick blue and red vectors are of the same size and have the same angle with the BD, the common side of the two triangles. Find congruent triangles.

    Last edited: Nov 12, 2016
  5. Nov 12, 2016 #4
    Sorry for taking so long to reply was kinda busy with some school work

    so congruent triangles are:
    BXD and BYD
    DPB and DRB
    QDB and SBD
    SPY and RQX
    ASP and QCR
    so that must mean that
    ABD and BCD are also congruent
  6. Nov 12, 2016 #5


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    What about the angles? Are DR and PB parallel?
    It is not enough, you have to point out which sides in ABD and in BCD correspond to each other, and these sides are parallel.
  7. Nov 12, 2016 #6
    DR an PB must be parallel since PD and BR are parallel and have the same length
    BS and QD are also parallel
    AD also has the same length as BC
    and AB has the same length as CD
    So AB=CD and AD=BC
    Therefore AB must be parallel to CD and AD must be parallel to BC
  8. Nov 12, 2016 #7


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    It is all right! But you should have said why the triangles are congruent (for example, BXD and BYD have two sides DY and BX equal, one side (BD) common, and the same angle between them. And it would have been enough to show that DR || PB involving DC || AB and SD || BQ, involving AD || BC.
  9. Nov 14, 2016 #8


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    Since this problem has been solved, I will show you a vector proof. I will start by assuming you know the medians intersect ##\frac 2 3## the way across a triangle. Then, using the same picture (reproduced here):
    Note that ##BX = \frac 2 3(BC + \frac 1 2 CD)## and ##YD = \frac 2 3(AD - \frac 1 2 AB)##. You are given that ##BX = YD## which implies ##BC + \frac 1 2 CD=AD - \frac 1 2 AB## or ##BC - AD = \frac 1 2 (-CD - AB)##. These are vectors, so we can write this as ##BC + DA = \frac 1 2 (DC + BA)##.
    But from the given quadrilateral we have ##AB + BC + CD + DA = \vec 0##. From this we get ##BC + DA = -AB -CD = BA + DC##. Substituting this in the left side of the last equation in the previous paragraph gives ##BA+DC =\frac 1 2(DC+BA)##, which implies ##DC+BA = \vec 0## or ##AB = DC##. So these opposite sides are equal and parallel.
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