In a parallelogram ABCD, the diagonal AC is 2 times the side AB and ∠ACD = 104

In a parallelogram ABCD, the diagonal AC is 2 times the side AB and ∠ACD = 104∘ Find the smaller angle between the diagonal of the parallelogram.

ABCD is a parallelogram.

AC = 2 * AB.

∠ACD = 104∘.

∠COD -?

According to the property of a parallelogram, its opposite sides are equal.
AB = CD, BC = DA.

According to the parallelogram property, the diagonals are halved by the intersection point.
AO = OC, BO = OD.

Consider an isosceles triangle Δ COD. He has ∠ACD = 104∘, which means that 180∘ – 104∘ = 76∘ remains at equal angles ∠COD and ∠ODC.
∠COD = ∠ODC = 76∘ / 2 = 38∘.

Answer: the smallest angle between the diagonals of a parallelogram is ∠COD = 38∘.