In the parallelogram ABCD, the bisector of angle A divides the side BC

In the parallelogram ABCD, the bisector of angle A divides the side BC into segments BK = 3 cm, KC = 2 cm. Find the perimeter of the parallelogram.

1. Since ∠BAK = ∠KAD (AK is a bisector), and ∠KAD = ∠AKB (lying crosswise), then △ ABK is isosceles.

2. It turns out that AB = BK = CD = 3 (cm);

3. BC = BK + KC = DA = 3 + 2 = 5 (cm);

4. Hence the perimeter of the parallelogram ABCD is:

P (ABCD) = AB + DC + CD + DA = 3 + 5 + 3 + 5 = 16 (cm);

5. Answer: P (ABCD) = 16 (cm).