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

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

1. The length of the side AB is equal to:
AB = AM + MB = 15 + 3 = 18.
Since the opposite sides of the parallelogram are equal, then AB = CD = 18.
2. Since DM is the bisector of ∠D, then ∠MDA = ∠MDC = ∠D / 2.
Since side AB is parallel to side CD, the bisector DM is a secant intersecting two parallel lines. Thus, ∠MDC = ∠AMD = ∠D / 2 as criss-crossing ⇒ ∠MDC = ∠AMD = ∠MDA.
In △ MAD ∠MDA = ∠AMD ⇒ △ MAD is isosceles, then AM = AD = 15.
3. Since ABCD is a parallelogram, then AD = BC = 15.
The perimeter of ABCD is:
P = AB + BC + CD + AD = 18 + 15 + 18 + 15 = 66.
Answer: P = 66.