Given is a parallelogram ABCD with ∠A<90o and ∣AB∣<∣BC∣. The angular bisector of angle A intersects side BC in M and intersects the extension of DC in N. Point O is the centre of the circle through M,C, and N. Prove that ∠OBC=∠ODC.
[asy]
unitsize (1.2 cm);pair A, B, C, D, M, N, O;A = (0,0);
B = (2,0);
D = (1,3);
C = B + D - A;
M = extension(A, incenter(A,B,D), B, C);
N = extension(A, incenter(A,B,D), D, C);
O = circumcenter(C,M,N);draw(D--A--B--C);
draw(interp(D,N,-0.1)--interp(D,N,1.1));
draw(A--interp(A,N,1.1));
draw(circumcircle(M,C,N));
label("∘", A + (0.45,0.15));
label("∘", A + (0.25,0.35));dot("A", A, SW);
dot("B", B, SE);
dot("C", C, dir(90));
dot("D", D, dir(90));
dot("M", M, SE);
dot("N", N, dir(90));
dot("O", O, SE);
[/asy] geometryparallelogramequal angles