# Given an isosceles trapezoid ABCD A perpendicular drawn from vertex B to the larger base AD

Given an isosceles trapezoid ABCD A perpendicular drawn from vertex B to the larger base AD divides this base into two segments, the largest of which is 7 cm, find the midline of the trapezoid.

Through point C we draw the height of the trapezoid CH.

The BCDM quadrangle is a parallelogram, since its opposite sides are parallel in pairs.

Then BC = HM.

The middle line of the trapezoid is: МР = (ВС + АD) / 2 = (НМ + АD) / 2.

AD = DH + AH, then MP = (HM + DH + AH) / 2.

Since the trapezoid is isosceles, the triangles ABН and CDM are equal in hypotenuse and acute angle, then DМ = АН.

MP = (HM + DH + DM) / 2.

НM + DM = DH, then MP = (DH + DH) / 2 = DH = 7 cm.

Answer: The height of the trapezoid is 7 cm.

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