In parallelogram ABCD, the straight line containing the bisector of angle B forms an angle of 40 °

In parallelogram ABCD, the straight line containing the bisector of angle B forms an angle of 40 ° with the straight line CD. Find angle A.

Since BM is the bisector of the angle ABC, then AB = AM, and the triangle ABM is isosceles.

Let us prove that triangles ABM and MDK are similar.

Angle AMB = KMD as vertical angles, angle BAM = MDK as criss-crossing angles at the intersection of parallel lines AB and СK secant ВK, then triangles ABM and MDK are similar in two angles.

Then AB / AM = DK / DM = 1, then DK = DM, and the MDK triangle is isosceles.

Then the angle DKM = DMC = 40, then the angle AMD = ABM = 40, and the angle BAD = 180 – 40 – 40 = 100.

Answer: Angle A is 100.