The obtuse angle bisector of a parallelogram divides the opposite side in a ratio of 1: 3

The obtuse angle bisector of a parallelogram divides the opposite side in a ratio of 1: 3, counting from the top of the acute angle. Find the larger side of the parallelogram if its perimeter is 60.

Parallelogram ABCD with obtuse angle B, larger base AD;
The bisector of angle B intersects side AD at point E and divides angle ABD into two equal angles;
We denote AE = X, then by condition ED = 3X, and AD = 4X;
In a parallelogram, the sum of the angles adjacent to one side is 180 °;
∡ BAD + ∡ ABC = 180 °;
In triangle ABE, the sum of the angles is also 180 °;
∡ BAD + ∡ ABC / 2 + ∡ BEA = 180 °;
Hence ∡ BEA = ∡ ABC, triangle ABC is isosceles, side AB = AE = X in it;
The perimeter of the parallelogram is (X + 4X) * 2 = 60, hence X = 6, and the base AD = 4X = 24.
Answer: 24