In parallelogram ABCD on side BC, point K is marked so that BK = AB. Find the BCD angle if the KAD angle is 20.

The angle KAD, by condition, is equal to 20, then the angle BKA is also equal to 200, as the angle KAD is crosswise.

Consider a triangle ABK, in which, by condition, AB = BK, therefore, the triangle is equilateral, which means that its angles, at the base, are equal. Angle ВAK = ВKA = 20.

Then the angle BAD, the parallelogram is 20 + 20 = 40. Since the angles of the opposite parallelogram are equal to each other, the angle BCD = BAD = 40.

Answer: ВСD = 40.