In trapezium ABCD, a straight line is drawn from the apex of angle B, parallel to side CD and intersecting side AD

In trapezium ABCD, a straight line is drawn from the apex of angle B, parallel to side CD and intersecting side AD at point K so that angle AKB is 65 degrees, angle A is 35 degrees. What is the BCD angle?

Since the straight line drawn from the angle B to the AD side is parallel to the CD side, since the upper and lower bases of the trapezoid are always parallel, we have a figure in which the opposite sides are respectively parallel to each other. This figure is a parallelogram.
The opposite angles of the parallelogram are equal.
Since the angle AKD is 180 °, we can find the angle BKD:
∠ВКD = ∠АКD – ∠ВКА = 180 ° – 65 ° = 115 °.
∠BCD = ∠ВКD and is also equal to 115 °.
ANSWER: ∠BCD = 115 °.