The bisectors of angles A and B of parallelogram ABCD meet at point N, which lies on side CD. Prove that N is the midpoint of CD.

Let the angle A = α, angle B = β.

By the property of the bisector of the parallelogram, the bisector of the angles of the parallelogram cut off the isosceles triangles from it.

This can be proved by considering the trapezoid BNDA, where the AB side is parallel to DN.

Then the angle N = 180º – ABN = 180º – β / 2.

The angle BNC as adjacent to BND is 180º – (180º – β / 2) = β / 2.

Similarly, one can prove that the angle AND = α / 2.

Then, triangles BCN and ADN are isosceles.

Therefore, BC = CN, AD = DN.

Since BC = AD in the parallelogram, it means that СN = DN.

So N is the middle of the CD.