The bisector of angle A of parallelogram ABCD intersects side BC in its middle M perimeter of triangle ABM

The bisector of angle A of parallelogram ABCD intersects side BC in its middle M perimeter of triangle ABM is 16 cm and length AM is 1 cm greater than side AB. Find the perimeter of the parallelogram.

Since AM is the bisector of the angle BAD, the triangle ABM is isosceles, AB = BM.

Let the length AB = BM = X cm, then, by condition, AM = (AB + 1).

The perimeter of the triangle ABM is equal to: Ravm = AB + BM + AM = X + X + X + 1 = 16 cm.

3 * X = 16 – 1 = 15 cm.

X = AB = BM = 15/3 = 5 cm.

Since point M is the middle of BC, then MC = BM = 5 cm, then BC = 2 * 5 = 10 cm.

Then Ravsd = 2 * (AB + BC) = 2 * (5 + 10) = 30 cm.

Answer: The perimeter of the parallelogram is 30 cm.