SOLUTION: ABCD is a rectangle with M the midpoint of BC. AC intersects MD at N. Find area of triangle NCD: area of ABMN

  

  

See the Figure on the right. 

Look in it attentively and identify all given lines and points. 

In addition to the given lines, I drew  the line BK with the 

point K as the middle point of the side AD. 

Let L be the intersection point of BK and AC.

Then it is clear from symmetry that |AL| = |CN| (congruent, have equal lengths).

Also from symmetry, it is clear that |AN| = |LC|.

   (One could find more complicated arguments, but it is enough for me now).

 
  


                    Figure. 

 
 
It implies that  |AL| = |LN| = |NC| = |AC|.   (Actually, it is well known property for this situation).
               (see the lesson Solved problems on Parallel lines cutting off congruent segments in transverse lines> in this site).


If so, then the height of the triangle DNC is exactly one third of the side |BC|.

Thus the area of the triangle DNC is one sixth () of the area of the rectangle.

Regarding the quadrilateral ABMN, one can show that its area is  =  of the area of the rectangle.