In trapezoid ABCD, angle A is straight, lateral side BA =. It is known that angle BCA = 60

In trapezoid ABCD, angle A is straight, lateral side BA =. It is known that angle BCA = 60 angle BDA = 30. Find the center line.

Since the trapezoid ABCD is rectangular, the triangles ABC and ABD are rectangular.

In a right-angled triangle ABC tg60 = AB / BC.

BC = AB / tg60 = √3 / √3 = 1 cm.

In a right-angled triangle ABD tg30 = AB / AD.

AD = AB / tg30 = √3 / (1 / √3) = √3 * √3 = 3 cm.

Let’s define the middle line of the trapezoid.

KM = (BC + AD) / 2 = (1 + 3) / 2 = 4/2 = 2 cm.

Answer: The middle line of the trapezoid is 2 cm.