One of the corners of the parallelogram ABCD is 5 times larger than the other, and the diagonal BD

One of the corners of the parallelogram ABCD is 5 times larger than the other, and the diagonal BD is the height, with D = 5cm. Find the length of the side of CD.

AB = CD, BC = AD.
The sum of all internal angles of the quadrilateral is 360 degrees, then in the parallelogram ABCD:
angle A + angle B + angle C + angle D = 360 degrees.
Angle A = angle C – let’s designate them as x, then angle B = angle D is 5x:
x + 5x + x + 5x = 360;
12x = 360;
x = 360/12;
x = 30.
Angle A = angle C = x = 30 degrees, angle B = angle D = 5x = 5 * 30 = 150 degrees.
It is known from the properties of a parallelogram that a parallelogram is divided by a diagonal into 2 equal triangles.
Consider a triangle ADB: angle BDB = 90 degrees (since BD is height), angle DAB (angle A) = 30 degrees, BD = 5 cm (by condition) – leg, AB – hypotenuse (since it lies opposite an angle equal to 90 degrees ). BD lies opposite the angle DAB, and from the properties of a right-angled triangle it is known that the leg, which lies opposite an angle of 30 degrees, is exactly 2 times less than the hypotenuse. Then:
AB = BD * 2;
AB = 5 * 2 = 10 (cm).
AB = CD = 10 cm.
Answer: CD = 10 cm.