# The bisector of an acute angle A of parallelogram ABCD intersects side BC at point M, which divides BC

The bisector of an acute angle A of parallelogram ABCD intersects side BC at point M, which divides BC into two segments 8 cm and 12 cm. Line AM intersects the continuation of side CD at point F. Find the length of segment DF.

Determine the length of the segment BC. BC = BM + CM = 12 + 8 = 20 cm.

The bisector AM cuts off the isosceles triangle ABM from the lateral side AB, then AB = BM = 12 cm.

The lengths of the opposite sides of the parallelogram are equal, then СD = AB = 12 cm, AD = BC = 20 cm.

Triangles ABM and MFC are similar in two angles, then AB / BM = FC / CM.

12/12 = FC / 8.

FC = 8 cm.

Then DF = 12 + 8 = 20 cm.

Answer: The length of the segment DF is 20 cm.

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