# Given a trapezoid ABCD. A straight line drawn from vertex B parallel to the side of CD

Given a trapezoid ABCD. A straight line drawn from vertex B parallel to the side of CD, intersects the larger base at point E. The perimeter of triangle ABE is 18 dm, and the segment ED = 5 dm. Find the perimeter of this trapezoid.

The BCDE quadrangle is a parallelogram, since its opposite sides are parallel. Then BC = CD, BC = ED = 5 cm.

The perimeter of the triangle Rave = ABE = AB + BE + AE = AB + CD + AE.

The perimeter of the trapezoid is equal to Ravsd = AB + BC + SD + AD = AB + BC + CD + AE + ED = (AB + CD + AE) + BC + DE = Pave + BC + ED = 18 + 5 + 5 = 28 cm.

Answer: The angles of the trapezoid are 60 and 120.

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