Problem 5. Answer: $B D=3$.

Option 0

In triangle $A B C$, the length of side $A C$ is $2 b$, side $A B-c$, and the angle at vertex $A$ is $60^{\circ}$. Point $M$, located on median $B E$, divides it in the ratio $B M: M E=p: q$. Line $A M$ intersects side $B C$ at point $D$. Find the length of segment $B D$.
Answer: $B D=\frac{p}{p+2 q} \cdot \sqrt{4 b^{2}+c^{2}-2 b c}$

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b90898f214fc60b7b26g-3.jpg?height=385&width=462&top_left_y=1709&top_left_x=794)

By the Law of Cosines, we find the length of side $B C: B C^{2}=4 b^{2}+c^{2}-2 b c$. Menelaus' Theorem:

![](https://cdn.mathpix.com/cropped/2024_05_06_5b90898f214fc60b7b26g-3.jpg?height=325&width=671&top_left_y=2199&top_left_x=772)

Triangle $A B C$ and the transversal $M P$ are arbitrary: $\frac{A M}{M B} \cdot \frac{B N}{N C} \cdot \frac{C P}{P A}=1$

Calculations:

$$
\begin{aligned}
& \frac{\sqrt{4 b^{2}+c^{2}-2 b c}-B D}{B D} \cdot \frac{p x}{q x} \cdot \frac{1}{2}=1 \rightarrow \sqrt{4 b^{2}+c^{2}-2 b c}-B D=\frac{2 q}{p} \cdot B D \rightarrow \\
& \rightarrow B D=\frac{p}{p+2 q} \cdot \sqrt{4 b^{2}+c^{2}-2 b c}
\end{aligned}
$$

## In Variant 1

$$
b=4, c=3, p=3, q=2 \rightarrow \sqrt{4 b^{2}+c^{2}-2 b c}=\sqrt{64+9-24}=7 \rightarrow B D=\frac{3}{3+4} \cdot 7=3
$$

## Variant 2