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Need help with a quadrilateral proof please!

  • #1

Homework Statement



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


Homework Equations





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?
 

Answers and Replies

  • #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.
 
  • #3
19
0
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:
  • #4
thank you alot. i didn't even notice those bigger traingles.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
249
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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