AB= 2/3 BC

Get BC= 12, AB=8.

Let the high AP of the parallelogram pass through point A to BC. In the right triangle ABP, the angle ABC = 60 and AB = 8°, so the high AP=4 and the root number 3.

2. Angle A = 150, so angle B = 30. If after A, AP is perpendicular to BC and BC intersects at point P, then the distance between AP and BC is in the right triangle ABP, with angle B = 30 and AB = 6°, so AP=3, that is, the distance between AD and BC is equal to 3.