# The bisector of angle A of the parallelogram ABCD intersects its side BC at point E.

The bisector of angle A of the parallelogram AВСD intersects its side BC at point E. find the area of the parallelogram if BE = 5 EC = 2 and the angle ABC = 150 °.

Since AE is the bisector of angle A, the angle BAE = DAE.
The BC side is parallel to AD, then the secant AE forms equal, criss-crossing angles EAD and BEA.
Then the angle BAE = BEA = DAE, and the triangle ABE is isosceles, AB = BE = 5 cm.
BC side length = (VK + SK) = 5 + 2 = 7 cm.
Then HELL = BC = 7 cm.
Let’s calculate the area of the parallelogram. S = AB * BC * SinABS = 5 * 7 * Sin150 = 35 * 1/2 = 17.5 cm2.
Answer: The area of the parallelogram is 17.5 cm2.

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