In trapeziums ABCD, the middle line PQ is drawn where side BC = 10cm, and AD = 14cm

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.