Point E is taken in trapezoid ABCD on the larger base AD. It is known that the angle ABC = 130 degrees, the angle

Point E is taken in trapezoid ABCD on the larger base AD. It is known that the angle ABC = 130 degrees, the angle BCE = 50 degrees. Prove that line segments AC and BE have a common midpoint.

According to the properties of a trapezoid, the sum of the angles at its lateral sides is 180.

Then the angle ВAD + ABC = 180.

Angle BAD = 180 – ABC = 180 – 130 = 50.

Consider a quadrilateral ABCE, the sum of the angles of which is 360.

Then the angle AEC = 360 – 130 – 50 – 50 = 130.

Angle ABC = AEC = 130.

Angle BAE = BCE = 50.

A quadrilateral whose opposite angles are pairwise equal is a parallelogram.

By the property of a parallelogram, its diagonal, at the point of intersection, is divided in half.

Then the point O is the midpoint of the segments AC and BE, as required.