Berkeley Math Circle

Solution:

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 $B M'$ symmetric to $B M$ with respect to the angle bisector of $A B C$ (point $M'$ is on the line $A C$). $B M'$ bisects the angle $D B E$. Using the same property of the angle bisector, we get $\frac{E M'}{B E} = \frac{D M'}{B D}$. Subtracting from this $\frac{C E}{B E} = \frac{C M}{B M}$ we get $\frac{C M'}{B E} = \frac{A M'}{B D}$ or $\frac{B D}{B E} = \frac{A M'}{C M'}$.

Now it remains only to find the ratio in which $M'$ 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

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