In a trapezoid, the distance between the midpoints of the bases is equal to the half difference

In a trapezoid, the distance between the midpoints of the bases is equal to the half difference of the lengths of the bases. Find the sum of the angles for a larger base.

Consider a trapezoid ABCD. Let points G, F be the midpoints of the bases AD, BC.

Let the point of intersection of lines AB and CB be point E.

Let’s connect point E to point F.

Segment EG is the median of triangle AED. Since BC is parallel to AD, EG crosses BC exactly in the middle. Obviously, the points E, F, G are collinear.

Draw from the vertex B the segment BH parallel to EG.

notice, that

AH = AG – HG = AG – BF = 1/2 * AD – 1/2 * BC = (AD – BC) / 2,

those. AH is equal to the half difference of the base lengths.

But according to the problem statement, BH = FG = (AD – BC) / 2.

Therefore, AH = BH and triangle ABH is isosceles and angles HAB = HBA.

Since EG is parallel to BH, then HBA = GEA and therefore, triangle AEG is also isosceles and GA = GE.

But GA = GD. Therefore GA = GD = GE. Therefore, G is the center of the circumscribed circle of triangle AED and AD is the diameter of this circle.

Hence it follows that the angle AED = 90 °. Then we have:

BAD + CDA = 180 ° – 90 ° = 90 °.

Answer: 90 °.