Connect AB so that O Q is parallel to AB through P, and MN EF crosses OQ respectively.
Because MN is parallel to EF AB, parallel to OQ, and AB is parallel to OQ, that is, parallelogram ABQO, and CD is the center line, AB is parallel to MN.
(It's too much trouble, briefly explain the proof process) Then take the center X of AB and connect it with the ox to prove that the quadrilateral AY(xpO is a parallelogram), and then take Y (the midpoint of Y(XP) to prove that the triangle ACE is all equal to CPY (there are many ways, I believe you will) and then the area is equal.
Similarly, the area below is equal.
Zheng bi