# In the parallelogram ABCD, the perimeter is 90, the bisector of the angle BAD divides the side BC in a ratio of 2: 5

In the parallelogram ABCD, the perimeter is 90, the bisector of the angle BAD divides the side BC in a ratio of 2: 5, counting from the vertex B. Find the larger side of the parallelogram.

Let us denote the bisector AK. Angle AKB = angle KAD as criss-crossing angles at the intersection of parallel lines AD and BC. Then angle BAK = angle AKB. Triangle ABK – isosceles, side AB = BK.

Determine how much one part of the ratio of the parties is equal to.
2 * 2 + 2 * (2 + 5) = 90;

4 + 2 * 7 = 90;

18 = 90;

1 part = 5.

Determine the larger side of the parallelogram.
(2 + 5) * 5 = 7 * 5 = 35.

Answer: the large side of the parallelogram is 35.

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