In trapezoid $A B C D$, the bases $A D=8$ and $B C=4$ are given. On the extension of side $B C$, a point $M$ is chosen such that the line $A M$ cuts off a triangle from the trapezoid, the area of which is four times smaller than the area of the trapezoid.

Find CM.
Find the ratio $C K: K D$, where $K$ is the intersection point of $A M$ and $C D$.

## Solution

Let $S=S_{A B C D}$. Suppose $K$ is the intersection point of line $A M$ with side $C D$. Then, by the condition, $S_{A K D}=S / 4$.

Since $A D=2 B C$, we have $S_{A C D}=2 S_{A B C}=2 S / 3, S_{A K C}=2 S / 3-S / 4=5 S / 12$.

Therefore, $C{ }^{\prime} /{ }_{K D}=S_{A K C / S_{A K D}}=5 / 3$.

From the similarity of triangles $A K D$ and $M K C$, it follows that $C M=A D \cdot{ }^{C K} /$ KD $=40 / 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-24.jpg?height=331&width=1081&top_left_y=2081&top_left_x=497)

## Answer

$40 / 3$

Submit a comment