Krdytkyu

Solved!
Thank you to those who answered my question. I managed to solve it while I was away. Will edit this question with my solution so that others can check.

Problem: Prove that the line segments joining the midpoints of opposite sides of a
quadrilateral bisect each other.

My work:

$$U + Z = W + V$$

Let the segment from $frac12U$ to $frac12 W$ be vector $AB$.

Let the segment from $frac12 Z$ to $frac12V$ be vector $CD$.

$$frac12(AB) = frac12U + frac12W + V$$

$$frac12(CD) = frac12Z + frac12V + U$$

^I'm not sure if the above will help me, but it seems like the hint that was given:

let the vertices be $4A,4B,4C,4D$ chosen to ensure $A+B+C+D=0$ so the lines joining opposite midpoints meet at $0=frac12(frac12(4A+4B)+frac12(4C+4D))$.

the midpoint of the line joining $4A$ to $4B$ is $2A+2B$, and likewise the opposite midpoint is $2C+2D$

since $0=(2A+2B)+(2C+2D)$ the origin is thus the midpoint of the either line joining the midpoints of a pair of opposite sides

Assuming that you can show that the midpoints are, indeed, as given in the hint, subtract the two expressions, to get

$$frac14(u-v+z -w)=frac14((u+z) - (v+w)) = 0,$$ the last equality being true since $u+z$ and $v+w$ are both the upper left vertex of your quadrilateral.

Watch the signs of your vectors.
-U/2+V+w/2=AB

Also note that is ALL of AB and not AB/2 as you have written.

Make life easier on yourself. Make the left side 2U and the top 2Z.
Make the bottom 2V and the right hand side 2W.
Now you dont have to bother with all those fractions.
You can keep the same orientations ie U and W point generally upward and V and Z toward the right. If you use the opposite sense of these vectors you MUST use a minus sign before the vector.

Now make the point of intersection the origin to also make life easy on yourself.
Let vector A point from the origin to the midpoint of the 2U vector.
Let vector B point from the origin to the midpoint of the 2W vector.
Let Vector C point from the origin to the midpoint of the 2Z vector.
Let Vector D point from the origin to the midpoint of the 2V vector.

2U+2Z=2V+2W so U+Z=V+W or equivalently U-V+W-Z=0

Try to get this last equation in terms of vectors A, B, C, and D so you can say something about these vectors.

Now look to the upper right.
W-Z=-B+C

In the lower left
U-V=A-D

-1 times the upper right equation plus 1 times the lower left equation yields

-(W-Z)+(U-V)=-(-B+C)+(A-D)

U+Z-W-V=0 so B-C+A-D=0

(A+B)=(C+D)

But the A and C vectors point in different directions.
B is just a some multiple of A but with the opposite direction.
D is a multiple of C but in the opposite direction from C.

The equation then reads...some multiple of the A vector equals some multiple of the C vector. This is true iff the multiple of A and the multiple of C are both zero.

A and B must be equal in magnitude but in opposite directions to add to yield the zero vector. The same type of thing is true with vectors C and D. They must be equal in magnitude and opposite directions.

The huge leap in this problem is realizing that for 2 independent vectors ie pointing in different directions to be equal to each other, they must be the zero vector.

Another helpful hint is if a problem has lines that cross, label that point as something which allows you to create vectors to and from that point.

If a problem has a line broken into 2 equal parts, label each part with the same vector ex. V and label the entire line 2V. If the line is broken into 3 equal parts, use V as the one third portion and 3V as the entire vector. This eliminates the problem of having to work with fractions and at times lets you find and equivalent for the vector V using other vectors in a drawing.

In this question your trying to say what you can about the diagonals. Before you write equations about the parts in question, say what you can about the remainder of the drawing. Those equations might help you cancel some things out of the later equations. In this diagram writing about the items not asked ie the outside of the diagram... was not much. It was one equation. Still, this was critical to finding a solution.