Problem

Source: Dutch NMO 2020 p3

Tags: geometry, parallelogram, equal angles



Given is a parallelogram $ABCD$ with $\angle A < 90^o$ 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 $\angle OBC = \angle ODC$. [asy][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("$\circ$", A + (0.45,0.15)); label("$\circ$", 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][/asy]