ABCD is a trapezoid in which the side AB is perpendicular to the base and the side

ABCD is a trapezoid in which the side AB is perpendicular to the base and the side CD is equal to the diagonal AC, BC = 1cm Find the midline of the trapezoid.

The ACD triangle is isosceles, since, by condition, AC = CD. Let us draw from the vertex C the height CH, which in the isosceles triangle will also be the median, then AH = DH.

Quadrangle ABCH is a rectangle, since BC is parallel to AD, and AB and CH are perpendicular to the bases. Then AH = BC = 1 cm.

AD = AH + DH = 1 + 1 = 2 cm.

Let’s define the middle line of the trapezoid.

KM = (BC + AD) / 2 = (1 + 2) / 2 = 1.5 cm.

Answer: The middle line of the trapezoid is 1.5 cm.