Let AD intersects BC at K. L is the midpoint of EK. The key is: L lies on AP ,CQ, MN,and LE^2=LK^2=LM*LN=LA*LP(1). Then we suppose that MN intersects BD at T',we only need to prove P,E,T',Q are concyclic,and you can find L also lies on this circle. It's easy to prove afterwards. proof of (1): It's well-known that M,N,L are collinear. ∠ENK+∠EMK=∠END+∠DNK+∠EMK=∠EMA+∠KMB+∠EMK=180°,so LE^2=LK^2=LM*LN. Suppose that EK intersects circle ABCD at X,Y, then K,E;X,Y are harmonic. The corollary is LE^2=LK^2=LX*LY=LM*LN,then it is easy to see A,P, L are collinear,as well as C,Q,L. （sorry for my poor English）

Newton line&DIT, very nice problem