In the parallelogram ABCD, the bisector AK is drawn, which intersects the BC at ВK = 5, KС = 2

In the parallelogram ABCD, the bisector AK is drawn, which intersects the BC at ВK = 5, KС = 2, find the Perimeter ABCD.

Since AK is a bisector, the angle BAK = DAK.

Angle AKB = DAK as criss-crossing angles at the intersection of parallel straight lines BC and AD secant AK.

Then the angle AKB = BAK, and then the triangle ABK is isosceles with the base AK, which means AB = BK = 5 cm.

Side length BC = BC + СK = 5 + 3 = 8 cm.

Then the perimeter of the parallelogram is: Ravsd = 2 * (AB + BC) = 2 * (5 + 8) = 26 cm.

Answer: The perimeter of the parallelogram is 26 cm.