In the parallelogram, point F is the midpoint of BC. Find the ratio of the area of triangle ABF to the area of quadrilateral AFCD.
Let’s draw a diagonal AC, which divides the parallelogram into two equal triangles.
Sас = Sасд.
In the triangle ABC, the segment AF is the median, since, by condition, BF = CF, then Sawf = Sacf.
Then Sacd = 2 * Savf.
Safcd = Sacd + Sacf = Sacd + Savf = 3 * Savf.
Then Savf / Safcd = 1/3.
Answer: The area ratio is 1/3.
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