Find the ratio of the bases of the trapezoid, given that its midline is divided into three equal parts by the diagonals.

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Let $AD=a, BC=b$ be the bases of trapezoid $ABCD (a>b)$, $M$ and $N$ be the midpoints of sides $AB$ and $CD$ respectively, and $K$ be the intersection point of the midline $MN$ with diagonal $AC$. Then $MK=1/2 BC=b/2$ as the midline of triangle $ABC$.

According to the condition, $MN=1/2(a+b)=3b/2$. From this, $a=2b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-01.jpg?height=311&width=592&top_left_y=805&top_left_x=733)

Answer

$1: 2$.

Problem