# In an isosceles trapezoid ABCD, the segment BF is parallel to CD and cuts off the rhombus BCDF.

**In an isosceles trapezoid ABCD, the segment BF is parallel to CD and cuts off the rhombus BCDF. The acute angle of the trapezoid is 60 degrees. Find the sides of the trapezoid if the perimeter is 20 cm.**

Since the trapezoid is equilateral, then AB = CD, and the segment BF cuts off the rhombus from the trapezoid, then B = BC = CD = FD = BF.

Consider a triangle ABF, in which AB = BF, hence the rectangle is isosceles, and the angles at the base AF are equal. Since the angle A of the trapezoid is 60, then the angle F of the triangle ABF is 60. Since the sum of the angles of the triangle is 180, then the angle B of the triangle ABF is 60, then the triangle ABF is equilateral, and AB = BF = AF.

Then AB + BC + CD + (AF + FD) = 20 cm.

AB + AB + AB + 2 * AB = 20.

5 * AB = 20.

AB = 20/5 = 4 cm.

BC = 4 cm.

CD = 4 cm.

AD = 2 * 4 = 8 cm.

Answer: AB = 4 cm, BC = 4 cm, CD = 4 cm, AD = 8 cm.