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Need help with a quadrilateral proof please!

  1. May 10, 2008 #1
    1. The problem statement, all variables and given/known data

    E 1\---------------------------------1N
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    K 1---------\-M---------------------1L
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    1 \ 1
    J 1---------------------------------\1U
    JUNE is a quadrilateral
    K is the midpoint of line JE
    L is the midpoint of line UN
    line KL and line UE bisect each other at point M
    Prove:JUNE i a parallelogram


    2. Relevant equations



    3. The attempt at a solution

    statment 1 reason
    --------------------------------------------------
    1)K is the midpoint of line 1 1) given
    JE
    2)line KE is congruent to 1 2) midpoint is the center of a line segment
    line KJ
    3)L is the midpoint of line 1 3) given
    UN
    4)line NL is congruent to line 1 4) midpoint is the center of a line segment
    LU
    5)line KL and line UE bisect 1 5) given
    each other at M
    6)line EM is congruent to line 1 6) line bisector splits the line segment in half
    MU
    7)line KM is congruent to line 1 7) line bisector splits the line segment in half
    ML
    8)Triangle EKM is congruent to1 8) theorem of Side,Side,Side
    triangle ULM
    9)line EK is congruent to line 1 9)corresponding parts of congruent triangles and congruent
    UL
    10)line KJ is congruent to line 1 10)substitution
    NL
    11)line EJ is congruent to line 1 11)substitution
    NU
    This is where i am stuck,how can I prove that either line EN is congruent to line JU or that line EJ is parallel to line NU?
     
  2. jcsd
  3. May 10, 2008 #2
    sorry the picture didn't come out the way it was suppose to. it is suppose to have a diagnol EU and line NU.
     
  4. May 10, 2008 #3
    I'm going to have [tex]\overline{XY}[/tex] denote the length of some line segment XY.

    let: [tex]a = \overline{KM} = \overline{ML}[/tex].

    By similar triangles (KEM and JEU), [tex]\overline{JU} = 2a[/tex].

    By similar triangles (MUL and EUN), [tex]\overline{EN} = 2a[/tex].

    It follows that [tex]\overline{JU} = \overline{EN}[/tex].



    let: [tex]b = \overline{UL} = \overline{NL}[/tex].

    By addition, [tex]\overline{NU} = 2b[/tex].

    By congruent triangles (UML and EMK), [tex]\overline{EK} = b[/tex].

    It follows that [tex]\overline{KJ} = \overline{EK} = b[/tex].

    By addition, [tex]\overline{EJ} = 2b[/tex].

    It follows that [tex]\overline{NU} = \overline{EJ}[/tex].



    [tex]\overline{JU} = \overline{EN}[/tex] and [tex]\overline{NU} = \overline{EJ}[/tex]; therefore, JUNE is a parallelogram.



    Also, don't worry about the picture too much, the text below it defines your problem perfectly.
     
    Last edited: May 10, 2008
  5. May 11, 2008 #4
    thank you alot. i didn't even notice those bigger traingles.
     
  6. May 11, 2008 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi shyguy10918! :smile:

    That's a proof using only lengths (and it's fine).

    You might like to try an alternative proof using angles, and the definition of a parallelogram as having parallel sides. :smile:
     
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