In trapezoid ABCD with bases BC and AD, angle BAD = 20 degrees, angle CDA = 70 degrees.

In trapezoid ABCD with bases BC and AD, angle BAD = 20 degrees, angle CDA = 70 degrees. Midline = 5, and the length of the line connecting the midpoints of the bases = 3. Find the base length AD.

Draw two straight lines from the middle of the smaller base (point K), KL – parallel AB and KM – parallel CD.

In the resulting triangle KLM, the angle KLM = 20, as the corresponding angle to the angle ABD at the intersection of parallel lines AB and KL. Based on the same angle KMO = 70. Since the sum of KLM + KMO = 90, then the angle LKM = 90, and the triangle KLM is rectangular.

The KO segment is the median drawn from the top of the rectangular corner, and is equal to half the length of the hypotenuse KO = LM / 2, then LM = 2 * KO = 2 * 3 = 6 cm.

Consider two quadrangles ABKL and KCDM, which are parallelograms since their opposite sides are parallel. Then BK = AL, and KC = MD.

Since point K is the middle of the segment BC, then BK = AL = KC = MD.

Let’s denote these segments by X, then BC = 2 * X, and AD = LM + 2 * X = 6 + 2 * X.

The middle line of a trapezoid is equal to the half-sum of the lengths of its bases, that is:

EF = (BC + AD) / 2.

5 = (2 * X + 6 + 2 * X) / 2.

10 = 4 * X – 6.

X = 1.

Then the length of the base AD = 6 + 1 + 1 = 8 cm.

Answer: AD = 8 cm.