The bisector of the angle A of the ABCD parallelogram intersects the BC side at point K. Find the perimeter

The bisector of the angle A of the ABCD parallelogram intersects the BC side at point K. Find the perimeter of this parallelogram if BC = 15 cm, KC = 9 cm.

BC is parallel to AD and the angles BKA and KAD are equal, as criss-crossing angles, because. ABCD-parallelogram
Triangle ABK is isosceles, (since the angles KAD and KAB are equal by the property of the bisector), therefore,
AB = BK = 15, CD is also 15, because. CD = AB
BC = BK = 9 + 15 = 24cm
BC = AD = 24cm
Hence it follows that
P (ABCD) = 2 * (15 + 24) = 78cm
Answer: 78 cm.