MathDB

Given is a parallelogram

ABCD

with

\angle A < 90^o

and

|AB| < |BC|

. The angular bisector of angle

A

intersects side

BC

M

and intersects the extension of

DC

N

. Point

O

is the centre of the circle through

M, C

, and

N

. Prove that

\angle OBC = \angle 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("

\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]

Problem Statement