The bisector of angle A of parallelogram ABCD makes an angle of 40 degrees with side BC. Find the angles of a parallelogram.

Consider the parallelogram ABCD
1) Mark the point of intersection of the bisector of angle A with side BC, as point H
It follows that the angle AHB = 40 (degrees)
2) angle HAD = angle AHB = 40 degrees (since the angles HAD and AHB are intersecting with parallel lines AD and BC, as well as secant AH)
3) angle BAH = angle HAD = 40 degrees (since, according to the condition of the problem, the bisector leaves the vertex A)
4) BAD angle = BAH angle + HAD angle
angle BAD = 40 + 40 = 80 (degrees)
5) angle ABC = 180 – angle BAD (since the angles in a parallelogram lying on either side add up to 180 degrees)
angle ABC = 180 – 80 = 100 (degrees)
Parallelogram angles:
ABC = 100 degrees
BAD = 80 degrees
Find the rest of the angles of the parallelogram:
angle CDA = angle ABC = 100 degrees (since opposite angles in a parallelogram are equal)
angle BCD = angle BAD = 80 degrees (since opposite angles in a parallelogram are equal)
Answer: 100; 80; one hundred; 80