HMMT February

Because $N L \| A C$ we have triangles $D X L$ and $D E C$ are similar. From angle chasing, we also have that triangle $D E C$ is similar to triangle $A B C$. We have $\angle X N A=180^{\circ}-\angle X N B=180^{\circ}-\angle L N B=180-C A B=\angle L M A$. In addition, we have $\frac{N X}{N A}=\frac{X D \cdot X E}{X L \cdot N A}=\frac{A B}{B C} \frac{X E}{L C} \frac{N M}{N A}=\frac{A B}{B C} \frac{E D}{D C} \frac{B C}{A B}=\frac{E D}{D C}=\frac{A B}{A C}=\frac{M L}{M A}$. These two statements mean that triangles $A N X$ and $A M L$ are similar, and $\angle X A B=\angle X A N=\angle L A M=\angle L A C$. Similarly, $\angle X A Y=\angle L A C$, making $A, X$, and $Y$ collinear, with $\angle Y A B=\angle X A B=\angle L A C$; ie. line $A X Y$ is a symmedian of triangle $A B C$. Then $\frac{Z B}{Z C}=\frac{A B}{A C} \frac{\sin \angle Z A B}{\sin \angle Z A C}=\frac{A B}{A C} \frac{\sin \angle L A C}{\sin \angle L A B}$, by the ratio lemma. But using the ratio lemma, $1=\frac{L B}{L C}=\frac{A B}{A C} \frac{\sin \angle L A B}{\sin \angle L A C}$, so $\frac{\sin \angle L A C}{\sin \angle L A B}=\frac{A B}{A C}$, so $\frac{Z B}{Z C}=\frac{A B^{2}}{A C^{2}}$.