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 theorm. 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.