## 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
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

## 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?

Related Precalculus Mathematics Homework Help News on Phys.org
sorry the picture didn't come out the way it was suppose to. it is suppose to have a diagnol EU and line NU.

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

tiny-tim