On the BC side of the ABCD parallelogram, point M is taken, so AB = BM, prove that AM is the bisector

On the BC side of the ABCD parallelogram, point M is taken, so AB = BM, prove that AM is the bisector of the BAD angle, find the perimeter of the parallelogram if CD = 8cm CM = 6cm.

1. Since the opposite sides of the parallelogram are pairwise parallel, AD∥BC ⇒ line AM is a secant intersecting two parallel lines.

Thus:

∠BMA = ∠MAD as criss-cross corners.

Since by the condition AB = BM, then △ ABM is isosceles, and ∠BMA = ∠BAM as the angles at the base of an isosceles triangle.

Thus:

∠BAM = ∠MAD ⇒ line AM divides ∠BAD in half, so it is the bisector of ∠BAD, which is what was required to prove.

1. By condition CD = 8 cm, CM = 6 cm.

Since the opposite sides of the parallelogram are equal, then:

AB = CD = 8 cm.

Since by the condition AB = BM, then BM = 8 cm.

Side BC is equal to:

BC = BM + CM = 8 + 6 = 14 (cm).

Since the opposite sides of the parallelogram are equal, then:

BC = AD = 14 cm.

1. The perimeter of the polygon is equal to the sum of the lengths of all its sides, then the perimeter of the parallelogram ABCD is:

P = AB + BC + CD + AD = 8 + 14 + 8 + 14 = 44 (cm).

Answer: P = 44 cm.