In trapezoid ABCD, the base of AD is 6 cm larger than the base of BC, and the middle line is 7 cm.

In trapezoid ABCD, the base of AD is 6 cm larger than the base of BC, and the middle line is 7 cm. Find the length of the segments into which the AC diagonal divides the middle line.

Let the length of the smaller base of the trapezoid be equal to X cm, then, by condition, the length of the base AD = (X + 6) cm.

According to the formula of the middle line of the trapezoid, KM = (AD + BC) / 2.

(BP + BC) = 2 * KM.

(X + 6 + X) = 2 * 7 = 14.

2 * X = 14 – 6 = 8.

X = BC = 8/2 = 4 cm.

AD = 4 + 6 = 10 cm.

In the triangle ABC, the segment OK is its middle line, then KO = BC / 2 = 4/2 = 2 cm.

Then OM = KM – KO = 7 – 2 = 5 cm.

Answer: The AC diagonal divides the midline into 2 cm and 5 cm segments.