# The middle line of the trapezoid is 4 m, the height is 3 m. Find the diagonals of the trapezoid

**The middle line of the trapezoid is 4 m, the height is 3 m. Find the diagonals of the trapezoid, considering that they are equal.**

Draw a straight line CK parallel to BD through the vertex C of the trapezoid.

The formed quadrilateral BCKD is a parallelogram, since its opposite sides are parallel. Then DK = BC, CK = ВK.

The segment AD = AD + DK = AD + BC, that is, it is equal to the sum of the lengths of the bases of the trapezoid.

Since MP = (BC + AD) / 2, then (BC + AD) = AK = 2 * MP = 2 * 4 = 8 m.

By condition, the diagonals of the trapezoid are equal to each other, which means that the triangle ACK is isosceles, AC = KC. The height of CH in an isosceles triangle is the median of the triangle, then AH = KH = AK / 2 = 8/2 = 4 m.

In a right-angled triangle ACН, leg CH = 3 m, leg AH = 4 m. Then triangle ACН is an “Egyptian” triangle and AC = CK = BD = 5 m.

Answer: The length of the diagonals of the trapezoid is 5 meters.