Similar Triangles

In triangle $ABC$, point $K$ on side $AB$ and point $M$ on side $AC$ are positioned such that $AK: KB = 3: 2$, and $AM: NC = 4: 5$.

Find the ratio in which the line passing through point $K$ parallel to side $BC$ divides the segment $BM$.
Draw a line through vertex $B$ parallel to $A C$.

## Solution

Let $F$ be the point of intersection of the specified line with side $A C$, and $P$ be the point of intersection of $B M$ and $K F$. Draw a line through vertex $B$ parallel to $A C$.

Let $T$ be the point of intersection of this line with line $K F$ (see figure). From the theorem of proportional segments, it follows that $F C = \frac{2}{5} A C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-32.jpg?height=360&width=740&top_left_y=1055&top_left_x=659)

Since $T B C F$ is a parallelogram, then $T B = F C = \frac{2}{5} A C$. From the similarity of triangles $T P B$ and $F P M$, it follows that $\frac{B P}{P M} = \frac{T B}{M F} = \frac{T B}{M C - F C} = \frac{\frac{2}{5} A C}{\frac{7}{45} A C} = \frac{18}{7}$

Answer

$18: 7$.