In an isosceles trapezoid $A B C D$, $A B=C D=3$, the base $A D=7, \angle B A D=60^{\circ}$. On the diagonal $B D$ there is a point $M$ such that $B M: M D=3: 5$.

Which side of the trapezoid, $B C$ or $C D$, does the extension of segment $A M$ intersect?
Let $K$ be the point of intersection of the lines $A M$ and $B C$. Compare $B K$ and $B C$.

## Solution

Let $P$ and $Q$ be the projections of vertices $B$ and $C$ onto the larger base $A D$ of trapezoid $A B C D$. From the right triangles $A P B$ and $D Q C$, we find that $A P = D Q = 3 / 2$. Therefore, $B C = P Q = 4$.

Let $K$ be the point of intersection of the lines $A M$ and $B C$. From the similarity of triangles $B M K$ and $D M A$, we find that $B K = B M^{A D} / M D = 21 / 5 = 4.2$.

Since $B K > B C$, the extension of segment $A M$ intersects side $C D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-22.jpg?height=343&width=692&top_left_y=669&top_left_x=686)

## Answer

$C D$.