ABCD trapezoid E belongs to AD, AC bisector angle ABC 130 ° angle BCE 50 ° prove that AC and BE have a common middle.

In a trapezoid, the sum of the angles at the lateral sides of the trapezoid is 180. Then the angle BAD = 180 – ABC = 180 – 130 = 50.

Consider a quadrangle ABCE, in which the opposite angles BAE and ALL are equal to each other, then the angle AEC = 360 – 130 – 50 – 50 = 130.

In a quadrilateral ABCE, opposite angles are pairwise equal, therefore, a quadrilateral is a parallelogram.

At the parallelogram, the diagonals, at the point of their intersection, are divided in half, therefore, AC and BE have a common midpoint at the point O. What was required to prove.