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Math Help - quadrilateral

  1. #1
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    quadrilateral

    a) Let ABCD be a quadrilateral. Consider the points and prove that EFGH is a parallelogram.

    b) Let (a,A), (b,B), (c,C), (d,D) be mass-points. Consider



    Under which conditions on a,b,c,d are the points E,F,G,H a parallelogram?
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  2. #2
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    Hint: A parallelogram is a quadrilateral whose diagonals bisect each other (so the midpoint of one diagonal is equal to the midpoint of the other diagonal).
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Hint: A parallelogram is a quadrilateral whose diagonals bisect each other (so the midpoint of one diagonal is equal to the midpoint of the other diagonal).
    I really still dont know how to do this problem. I drew an arbitrary quadrilateral and used the formula A+C=B+D for a parallelogram but still cant prove it.
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  4. #4
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    Hello, bigb!

    a) Let ABCD be a quadrilateral.

    Consider the points: . E \:=\:\tfrac{1}{3}(A + 2B),\;F \:=\:\tfrac{1}{3}(2B + C),\;G \:=\:\tfrac{1}{3}(C + 2D),\;H \:=\:\tfrac{1}{3}(2D+A)

    Prove that EFGH is a parallelogram.

    I must assume that those equations refer to the lengths of line segments,
    and that the sides are divided in the ratio 1:2 or 2:1.

    The diagram looks somthing like this:
    Code:
                A           E           B
                o - - - - - o - - - - - o
               *        *       *         *
              *     *               *       *
             *  *                       *     *
          H o                               *   *
           *    *                               * *
          *         *                               o F
         *              *                       *     *
        *                   *               *           *
       *                        *       *                 *
      o - - - - - - - - - - - - - - o - - - - - - - - - - - o
      D                             G                       C
    Draw diagonal BD.


    Consider \Delta AEH\text{ and }\Delta ABD
    . . \angle EAH \,=\,\angle BAD
    . . AE:AB \,=\,1:3 \quad\Rightarrow\quad AE \,=\,\tfrac{1}{3}AB
    . . AH:AD \,=\,1:3 \quad\Rightarrow\quad AH \,=\,\tfrac{1}{3}AD
    Hence: . \Delta AEH \sim \Delta ABD
    . . HE:DB \,=\,1:3 \quad\Rightarrow\quad HE \,=\,\tfrac{1}{3}DB .[1]
    . . \angle AHE\,=\,\angle ADB \quad\Rightarrow\quad HE \parallel DB .[2]


    Consider \Delta CFG\text{ and }\Delta CBD
    . . \angle FCG \,=\,\angle BCD
    . . CF:CB \,=\,1:3 \quad\Rightarrow\quad CF \,=\,\tfrac{1}{3}CB
    . . CG:CD \,=\,1:3 \quad\Rightarrow\quad CG \,=\,\tfrac{1}{3}CD
    Hence: . \Delta CFG \sim \Delta CBD
    . . FG:DB \,=\,1:3\quad\Rightarrow\quad FG \,=\,\tfrac{1}{3}DB .[3]
    . . \angle FGC \,=\,\angle BDC \quad\Rightarrow\quad FG \parallel DB .[4]


    From [1] and [3]: . HE\,=\,FG

    From [2] and [4]: . HE \parallel FG


    Theorem: If two sides of a quadrilateral are equal and parallel,
    . . . . . . . the quadrilateral is a parallelogram.


    Therefore, EFGH is a parallelogram.

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