1. May 10, 2008

### shyguy10918

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
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. May 10, 2008

### shyguy10918

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. May 10, 2008

### misho

I'm going to have $$\overline{XY}$$ denote the length of some line segment XY.

let: $$a = \overline{KM} = \overline{ML}$$.

By similar triangles (KEM and JEU), $$\overline{JU} = 2a$$.

By similar triangles (MUL and EUN), $$\overline{EN} = 2a$$.

It follows that $$\overline{JU} = \overline{EN}$$.

let: $$b = \overline{UL} = \overline{NL}$$.

By addition, $$\overline{NU} = 2b$$.

By congruent triangles (UML and EMK), $$\overline{EK} = b$$.

It follows that $$\overline{KJ} = \overline{EK} = b$$.

By addition, $$\overline{EJ} = 2b$$.

It follows that $$\overline{NU} = \overline{EJ}$$.

$$\overline{JU} = \overline{EN}$$ and $$\overline{NU} = \overline{EJ}$$; 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
4. May 11, 2008

### shyguy10918

thank you alot. i didn't even notice those bigger traingles.

5. May 11, 2008

### tiny-tim

Hi shyguy10918!

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.