The bisector of an obtuse angle of a parallelogram divides the opposite side in a ratio of 5: 8, counting

The bisector of an obtuse angle of a parallelogram divides the opposite side in a ratio of 5: 8, counting from the apex of the acute angle. Find the larger side of the parallelogram if its perimeter is 72.

Given a parallelogram ABCD, a bisector BK is drawn from an obtuse angle ABC, which divides the AD side into parts AK and KD, according to the condition AK: KD = 5: 8;

Since BK is a bisector, then ∠ ABK = ∠ KBC;

∠ AKB = ∠ KBC as criss-crossing angles at two parallel sides of the parallelogram AD and BC and secant BK;

This implies that ∠ ABK = ∠ BKA, so triangle ABK is isosceles and side AB = AK;

If side AD contains 5 + 8 = 13 parts, AB – 5 parts, then the perimeter of the parallelogram is

(13 + 5) * 2 = 36 parts;

By condition, the perimeter is 72, which means that the large side of the parallelogram is 72/36 * 13 = 26.