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Math Help - Wittenbauer's Parallelogram

  1. #1
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    Wittenbauer's Parallelogram

    Guys I need help and the internet is failing me.

    I need to prove Wittenbauer's Parallelogram. This is what it looks like: Wittenbauer's Parallelogram from Interactive Mathematics Miscellany and Puzzles

    Basically, any convex quadrilateral ABCD, if you trisect it's sides at S,T; U,V; W,X; Y,Z then TU, VW, XY, ZS form a parallelogram.

    All I really have to work with is area and similarity and I'm just completely stumped. I've been staring at it for a while and just am not seeing what initial connection I have to make in order to see the proof.



    Once I prove it's a parallelogram I have other stuff to prove and am hoping they follow easily afterward. Here is the other stuff:

    Prove that |TU| = 2/3 |AC| and |VX| = 2/3 |BD|

    and

    If ABCD is a rhombus, prove that the figure formed by TU, VW, XY, ZS is a rectangle and that it's area is equal to 8/9 |ABCD|.
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  2. #2
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    The answer is in the link you gave to the diagram

    The sides of the parallelogram are parallel to the diagonals of ABCD (which in fact explains why there's a parallelogram in the first place)
    Triangle TBU is similar to ABC, implying TU is parallel to AC. Do the same in the other corners and you have a quadrilateral with opposite sides parallel. Good luck with the rest of the questions.
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  3. #3
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    Re: Wittenbauer's Parallelogram

    But this same page explains that the lines drawn are parallel to the diagonals of the given quadrilateral. Since there are only two diagonals, the lines come in parallel pairs and thus form a parallelogram.
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  4. #4
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    Not sure how I missed that

    But ok, how do I prove that the sides are parallel to the diagonals?
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  5. #5
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    why parallel?

    Look at the triangle ACD, say, with points S on CD and T on AD such that S divides CD in ratio 2:1 while T divides AD in ratio 2:1. Then triangles STD and CAD are similar (having the same angle at D and proportional sides.) Thus all their angles are pairwise equal. In particular, say, angles at T and A are equal. But these are corrsponding angles for the lines ST and AC and a transversal AD. Their equality makes the lines ST and AC parallel.
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  6. #6
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    Thank you so much, I can't believe it too me that long to see it. Except the ratio is 1:3 isn't it? Well maybe it's just the way I'm thinking it.

    Anyone have hints on the other two problems?
    Last edited by HeirToPendragon; January 25th 2009 at 01:39 PM.
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  7. #7
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    Except the ratio is 1:3 isn't it? Well maybe it's just the way I'm thinking it.
    The ratio is indeed 1:3. Perhaps the question should have said the length of the side of the parallelogram passing through T and U instead of |TU|.
    Attached Thumbnails Attached Thumbnails Wittenbauer's Parallelogram-wtri.jpg  
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  8. #8
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    Heh, I didn't word it.

    Now the only thing I'm stuck on is, if ABCD is a rhombus, prove A'B'C'D' is a rectangle.

    I'm so close, but I keep using things that I don't think I can prove
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  9. #9
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    The diagonals of a rhombus intersect at right angles. The sides of A'B'C'D' are parallel to the diagonals, so just use corresponding angles.
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  10. #10
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    Can a rectangle be defined simply as a quadrilateral with four 90 degree angles?

    I've been sitting here for hours trying to figure out how I can show that CD is parallel to B'D' to show that |A'C'| = |B'D'|.

    Am I just thinking to hard?



    Wow, yes I am. There it is right there. Corollary: A parallelogram with at least one vertex angle a right angle is a rectangle. Damn I hate it when I think to hard.

    Thank you so much for smacking me
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