In a rectangular trapezoid ABCD, the height AB is equal to the sum of the bases AD and BC.

In a rectangular trapezoid ABCD, the height AB is equal to the sum of the bases AD and BC. The bisector of the angle ABC intersects the CD side at point K. In what relation does this point divide CD?

On line AB we mark point M so that BC = BM. Then the BCM triangle is isosceles and rectangular.

AB = AM + BM = AM + BC, and since, by condition, AB = BC + AD, then AM + BC = BC + AD, therefore AM = AD. Then the AMD triangle is rectangular and isosceles. Angle ВМС = AMD = 45, then the angle СMD = 180 – 45 – 45 = 90. The triangle СMD is rectangular.

ВM is the bisector of angle B, and therefore it is the height and median of the BCM triangle, then OM = OС. Then the СOК triangle is also rectangular. Rectangular triangles MD and СOК are similar in acute angle C, and since OM = OC, then CM = 2 * OC, then the coefficient of similarity of triangles is: K = OC / CM = 1/2, then СD / CK = 1/2.

Point K divides the СD segment in half.

Answer: Point K divides the segment СD in half.