In parallelogram ABCD, the bisector of an acute angle A intersects side BC at point N

In parallelogram ABCD, the bisector of an acute angle A intersects side BC at point N, BN = 5cm, NC = 3cm. find the perimeter of the parallelogram

Consider the resulting triangle ABN and the angles obtained at the intersection of lines AN and BC: <BAN; <BNA; <DNA.

Consider a triangle ABN, in which <BAN = <BNA, since <DNA = <BNA, as angles lying internally with parallel lines BC ∥ AD.
But <BAN = <DNA, since AN is the bisector of the angle <BAD by condition, so <BAN = <BNA.

Triangle ABN is isosceles, since we proved that the angles at the base of AN are equal. In an isosceles triangle, the sides are equal: AB = BN = 5 (cm).

BC = BN + NC = (5 + 3) = 8 (cm).
Perimeter ABCD = 2 * (AD + AB) = 2 * (8 + 5) = 26 (cm).