In trapeziums ABCD, the middle line PQ is drawn where side BC = 10cm, and AD = 14cm. Find the middle line of the trapezoid.
Draw a straight line through points B and Q up to the intersection with the extension of the base AD.
Triangles BCQ and KDQ are equal in side and adjacent angles CQ = DQ, angle BCQ = KQD as vertical angles, angle CDQ KQD as criss-crossing angles.
Then DK = BC = 10 cm, and AK = DK + ADD = 10 + 14 = 24 cm.
Since in triangle ABK point P is the middle of AB, and PQ is parallel to AK, then PQ is the middle line of triangle ABK, then PQ = AK / 2 = 24/2 = 12 cm.
Answer: The length of the middle line is 12 cm.
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