\section*{Problem 1 - 051021}

Let \(E\) be the midpoint of the diagonal \(DB\) of the parallelogram \(ABCD\). Point \(F\) is the point on \(AD\) such that \(|DA|:|DF|=3:1\).

How does the measure of the area of triangle \(\triangle DFE\) compare to that of quadrilateral \(ABEF\), assuming the same units of measurement are used?
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1221.jpg?height=285&width=534&top_left_y=603&top_left_x=264)

The heights of triangles \(D A B\) and \(D F E\) on \(g_{D B}\) and \(g_{D E}\) are parallel to each other because the sides \(D B\) and \(D E\) of the mentioned triangles lie on the same line. According to the 2nd ray theorem, the lengths of these heights are in the ratio of \(3:1\) due to the given condition. Furthermore, the lengths of the base sides \(D B\) and \(D E\) are in the ratio of 2:1.

Therefore, the areas of the two triangles are in the ratio of 6:1. Consequently, the area of triangle \(D F E\) to that of quadrilateral \(A B E F\) is in the ratio of 1:5.