In the triangle $A B C$ the angle $B$ is not a right angle, and $A B: B C=k$. Let $M$ be the midpoint of $A C$. The lines symmetric to $B M$ with respect to $A B$ and $B C$ intersect $A C$ at $D$ and $E$. Find $B D: B E$.
As $B C$ is the angle bisector in the triangle $M B E$, we have $\frac{C E}{B E}=\frac{C M}{B M}$ (by a well-known property of the angle bisector). Similarly, $\frac{A D}{B D}=\frac{A M}{B M}$. Draw a line $B M^{\prime}$ symmetric to $B M$ with respect to the angle bisector of $A B C$ (point $M^{\prime}$ is on the line $A C$ ). $B M^{\prime}$ bisects the angle $D B E$. Using the same property of the angle bisector, we get $\frac{E M^{\prime}}{B E}=\frac{D M^{\prime}}{B D}$. Subtracting from this $\frac{C E}{B E}=\frac{C M}{B M}$ we get $\frac{C M^{\prime}}{B E}=\frac{A M^{\prime}}{B D}$ or $\frac{B D}{B E}=\frac{A M^{\prime}}{C M^{\prime}}$.

Now it remains only to find the ratio in which $M^{\prime}$ divides $A C$. To do that, note that $M B C$ and $M B A$ have equal areas: $\frac{1}{2} B M \cdot B C \cdot \sin \angle M B C=$ $\frac{1}{2} B M \cdot B A \cdot \sin \angle M B A$. Therefore $\frac{\sin \angle M B C}{\sin \angle M B A}=\frac{A B}{B C}=k$. Hence

$$
\begin{aligned}
\frac{A M^{\prime}}{C M^{\prime}}=\frac{S_{A B M^{\prime}}}{S_{B C M^{\prime}}} & =\frac{\frac{1}{2} B M^{\prime} \cdot B A \cdot \sin \angle M^{\prime} B A}{\frac{1}{2} B M^{\prime} \cdot B C \cdot \sin \angle M^{\prime} B C} \\
& =k \cdot \frac{\sin \angle M^{\prime} B A}{\sin \angle M^{\prime} B C}=\frac{\sin \angle M B C}{\sin \angle M B A}=k \cdot k=k^{2}
\end{aligned}
$$