The bisector of angle A of parallelogram ABCD intersects side BC at point P with BP = PC.

The bisector of angle A of parallelogram ABCD intersects side BC at point P with BP = PC. find the sides of the parallelogram if its perimeter is 54 cm.

Angle DAP = BPA, as these are criss-cross corners. And since BAP = BPA, it follows that the triangle BAP is isosceles: AB = BP = PC.
BC = BP + PC, hence BC = 2AB
Let AB = x, then BC = 2x
The perimeter is: 2AB + 2BC = 2x + 2 (2x) = 2x + 4x = 6x
6x = 54
x = 9
AB = 9
BC = 2 * 9 = 18
Since the opposite sides in the parallelogram are equal, then AB = CD = 9, BC = AD = 18
Answer: AB = 9, CD = 9, BC = 18, AD = 18