The AC diagonal of the trapezoid ABCD divides the trapezoid into two triangles ABC and ACD.
Consider a triangle ABC. Since point K is the middle of AB, and KO is parallel to BC, then KO is the middle line of the triangle.
The length of the median line of the triangle is half the length of the base parallel to the midline. KO = BC / 2 = 5/2 = 2.5 cm.
Consider triangle ACD. Since point M is the middle of CD, and MO is parallel to AD, then MO is the middle line of triangle ACD.
Then OM = AD / 2 = 9/2 = 4.5 cm.
Answer: The length of the segments is 2.5 cm and 4.5 cm.
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