# The bisector of angle A and angle D of parallelogram ABCD intersect at a point belonging to BC

The bisector of angle A and angle D of parallelogram ABCD intersect at a point belonging to BC, perimeter ABCD is 46cm. Find the sides of a parallelogram.

Since AM and DM are the bisectors of the angles at the base of AD, they form isosceles triangles ABM and CDM.

Then AB = BM, CD = CM, and since AB = CD as opposite sides of the parallelogram, then BM = CM.

Let the length AB = X cm, then BM = CM = CD = X cm, and BC = AD = 2 * BM = 2 * X cm.

The perimeter of the parallelogram is: Ravsd = 2 * (AB + AD) = 2 * (X + 2 * X) = 6 * X = 46 cm.

X = 46/6 = 23/3 = 7 (2/3) cm.

AB = CD = 7 (2/3) cm.

BC = AD = 15 (1/3) cm.

Answer: The sides of the parallelogram are 7 (2/3) cm, 15 (1/3) cm.

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