Proving that MN is Parallel to AB and CD

[Styx]

Lines AB and CD are parallel. You are given M and N as midpoints of AD and BC, respectively.

Prove that MN is parallel to AB and CD

Ok, so I think I way (WAY) over complicated this. Can someone please suggest a shorter route to my answer? The worst part is that even after working out this own mess below I think my proof is half baked at best...

Work

Draw a perpendicular line from the midpoint of AB to line CD.
Draw a perpendicular line from the midpoint of CD to line AB.
The two lines overlap
Therefore, angles AQP, BQP, DPQ, CPQ are 90 degrees.

point Q is what I gave the midpoint of AB, P is the midpoint of DP and Point O is where AC and DB intercept

Draw a line from point B to point D, and from point A to Point C.
Consider triangles AQO and BQO
AQ = BQ since Q is at the midpoint between AB
angle AQO and angle BQO are both right angles
QO = QO (common to both triangles)
Therefore, triangles AQO and BQO are congruent as per the SAS condition of the triangle congruence theorem.

The same consideration can be applied to DPO and CPO to show that they are congruent as per the SAS condition of a triangle.

180 degrees - angle AOB = angle DOA
180 degrees - angle AOB = angle COB
Therefore, angle DOA = angle COB

AO = BO, DO = CO
Therfore, Triangles COB and DOA are congruent as per the SAS condition of a triangle.

AD = BC, AM = MD, BN = NC
Therefore, AM = MD = BN = NC

MN can only meet this conditions if it is a parallel line.
Therefore, MN is parallel to AB and DC

I am kind of unclear as to what I have to prove and what I can take for granted when I am working on these so I attempt to prove everything under the sun.

 

[Styx]

Ok, I just used Geometer's Sketchpad (great program btw) which disproved my first assumption that angles AQP, BQP, DPQ, CPQ are 90 degrees.

So, I guess I will start over...

Part B) States: Prove that MN =(1/2)(AB+CD) which I know is true as per Sketch pad. Can I use that in solving part a) or is that considered bad form?

The only thing I know for sure (other that what was given is that angle ABD = angle BDC, angle BAC = angle ACD

 

[Styx]

I rewrote my proof as:

AB and CD are parallel and M and N are the midpoints of AD and BC (given hypothesis).

Draw a perpendicular line from point B to line DC.
Draw a perpendicular line from point A to line DC.
Label them points O and P.
Since AO and BP are both perpendicular lines between the same two parallel lines AO = BP

Mark the midpoint of AO and BP, label them Q and R
Both points Q and R fall on line MN
Therefore, MN is parallel to AB and CD