# In parallelogram ABCD, angle B is equal to 120 degrees and the bisector of this angle divides side AD

In parallelogram ABCD, angle B is equal to 120 degrees and the bisector of this angle divides side AD into segments AE = 6 and BC = 2. Find the perimeter of the parallelogram. Determine the view of the quadrilateral ABCD and find its perimeter.

By the property of parallelogram angles, angle A = angle B (internal one-sided at the intersection of parallel straight lines of the third)

Let us find the angle A, if the angle B = 120 °, according to the condition.

Angle A + angle B = 180 °.

Angle A = 180 ° – Angle B = 180 ° – 120 ° = 60 °.

BE – bisector of angle B, halves it.

Angle ABE = 120 °: 2 = 60 °.

Consider triangle ABE, the sum of the angles is 180 °.

Angle A + angle ABE + angle BEA = 180 °.

60о + 60о + angle BEA = 180о.

angle BEA = 180 ° – 120 ° = 60 °.

All angles are equal, triangle ABE is regular, equilateral.

AB = BE = AE = 6 (by condition)

AD = AE + ED = 6 + 2 = 8.

P ABCD = 2 * (AB + AD).

P = 2 * (6 + 8) = 2 * 14 = 28.

Answer: P ABCD = 28, ABCD – – parallelogram.

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