In the parallelogram ABCD, the diagonal AC is 2 times the side of CD and the angle ACD = 72.

In the parallelogram ABCD, the diagonal AC is 2 times the side of CD and the angle ACD = 72. Find the acute angle between the diagonals of the parallelogram.

Since, by condition, the diagonal AC is twice as large as the side AB, and the diagonals of the parallelogram, at the point of intersection, are divided in half, then AO = CO = AB = AC / 2 cm.

In a parallelogram, the opposite sides are equal, then CD = AB = AC / 2.

Then, in the triangle CDO, CD = OC, and therefore the triangle COD is isosceles with the base OD, then the angle COD = DOC = (180 – 72) / 2 = 108/2 = 54.

Answer: The acute angle of intersection of the diagonals is 54.