In parallelogram ABCD, the bisector of angle D intersects side AB at point M, AM is 15cm, MB

In parallelogram ABCD, the bisector of angle D intersects side AB at point M, AM is 15cm, MB is 3cm Find the perimeter of ABCD.

∠MAD + ∠CDA = 180 ° since these are angles adjacent to one side of the parallelogram, hence

∠MAD = 180 ° – ∠CDA;

∠MDA = ∠CDA / 2, since the bisector bisects the angle;

The sum of the angles of the AMD triangle is 180 °;

∠MAD + ∠MDA + ∠AMD = 180 ° – ∠CDA + ∠CDA / 2 + ∠AMD = 180 °;

Hence, ∠AMD = ∠CDA / 2;

Since the angles ∠AMD = ∠MDA, the triangle AMD is isosceles with base MD, in which the sides AD and AM are equal.

AD = 15 cm, AB = 15 + 3 = 18;

The perimeter of the parallelogram is (15 + 18) * 2 = 66 cm.