The question's title and context don't match. The midpoints of AD and DC are not the midpoints of LO and MN, and how can the midpoint of MN be N when N is an endpoint? Also, L and O haven't been defined in the question. The answer below assumes M to be the midpoint of AD and N to be the midpoint of BC.
Extend the sides DA and CB so that they meet at O to form a triangle ODC. We need the length of the base DC (CD). Let OA=x and AM=MD=a. From similar triangles OBA and ONM we have OA/AB=OM/MN=x/12=(x+a)/14.
14x=12x+12a; 2x=12a so x=6a. OA/OC=AB/CD; x/(x+2a)=12/CD; 6a/8a=12/CD. 12/CD=3/4, CD=12*4/3=16.
OK. So that was too easy? Let's assume something different and see how far we get. This time, M is the midpoint of AD, as before, but N is the midpoint of DC. So CN=ND.
With N as the centre of a circle radius 14 a circle can be drawn. M is any point on the circumference. Join this point to D to form DM and extend this line to A such that DM=MA. The line AB parallel to DC is length 12. Join B to C. Many such arbitrary points M can be made, so many trapeziums ABCD can be drawn. Therefore, the length of DC cannot be determined because there are insufficient constraints between the length of AB and the length of DC, because the positioning of A and B are arbitrary.