The height of trapezoid $A B C D$ is 7, the bases $A D$ and $B C$ are 8 and 6 respectively. A line through point $E$, lying on side $C D$, intersects diagonal $A C$ at point $O$ in the ratio $A O: O C=3: 2$. Find the area of triangle $O E C$.

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$S_{O E C}={ }^{C E} /{ }_{C D} \cdot C O / C A S_{A C D}$.

## Solution

Let's extend the segment $B E$ to intersect the line $A D$ at point $M$. From the similarity of triangles $B O C$ and $M O A$, it follows that $B C: A M = C O: O A$. Therefore,

$A M = B C \cdot O A / O C = 9, D M = A M - A D = 1$.

Thus, the similarity ratio of triangles $B C E$ and $M D E$ is ${ }^{B C / / D M} = 6$. This means, ${ }^{C E} / D E = 6, { }^{C E} / C D = 6 / 7$. Consequently, $S_{O E C} = 6 / 7 \cdot 2 / 5 S_{A C D} = 6 / 35 \cdot 8 \cdot 7 = 48 / 5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-22.jpg?height=383&width=834&top_left_y=786&top_left_x=612)

## Answer

$9.6$.