[ Two pairs of similar triangles]

Points $A_{1}$ and $B_{1}$ divide the sides $B C$ and $A C$ of triangle $A B C$ in the ratios: $B A_{1}: A_{1} C=1: p$ and $A B_{1}: B_{1} C=1: q$. In what ratio does the segment $A A_{1}$ divide the segment $B B_{1}$?
Continue $B B_{1}$ until it intersects with the line passing through vertex $A$ parallel to $B C$.

## Solution

Let $M$ be the intersection point of segments $A A_{1}$ and $B B_{1}$. Draw a line through vertex $A$ parallel to $B C$, and extend $B B_{1}$ until it intersects this line at point $K$. From the similarity of triangles $A B_{1} K$ and $C B_{1} B$, it follows that ${ }^{B C} /_{A K}=B_{1} C /_{A B_{1}}=q$, and from the similarity of triangles $A M K$ and $A_{1} M B$, we have $A M / M A_{1}={ }^{A M} / M A_{1}={ }^{A K} / A_{1} B={ }^{B C} / A_{A_{1} B} \cdot A K / B C=p+1 / q$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9bfea73962495faf8ed9g-11.jpg?height=394&width=688&top_left_y=1884&top_left_x=685)

Answer

$p+1 / q$.