7.1.4 ** In $\square A B C D$, $A B \neq B C$, it is known that the ratio of the lengths of the diagonals $\frac{A C}{B D}=k$, ray $A M$ and ray $A D$ are symmetric about line $A C$, ray $B M$ and ray $B C$ are symmetric about line $B D$, $M$ is the intersection of ray $A M$ and $B M$, find $\frac{A M}{B M}$ (expressed in terms of $k$).
There are two possible scenarios: $A B>A D$ and $A B<A D$. There are differences in the diagrams and solutions, but here we only discuss the first scenario (the rest is left to the reader). Connect $M$ and the center $O$ of the parallelogram. We will prove $\triangle A O D \sim \triangle A M O, \triangle B O C \sim \triangle B M O$. From the given equations $\angle M A O=\angle D A O, \angle M B O=\angle C B O$, let $\angle D A O=\alpha, \angle C B O=\beta$, then $\angle A D O=\beta, \angle A O B=\alpha+\beta, \angle A M B=360^{\circ}-\angle M A O-\angle M B O-\angle A O B=$ $360^{\circ}-2 \alpha-2 \beta$. At this point, point $O$ is equidistant from the lines $A D$ and $A M, B C$ and $B M$. This indicates that point $\odot$ is equidistant from the lines $A M$ and $B M$, meaning $M O$ is the angle bisector of $\angle A M B$. Therefore, $\angle A M O=\frac{1}{2} \angle A M B=180^{\circ}-$ $\alpha-\beta=\angle A O D$, similarly $\angle B M O=\angle B O C$, thus $\triangle A O D \sim \triangle A M O, \triangle B O C \sim$ $\triangle B M O$. This leads to $\frac{A M}{A O}=\frac{A O}{A D}, \frac{B M}{B O}=\frac{B O}{B C}$, hence $\frac{A M}{B M}=\frac{A O^{2}}{B O^{2}}=k^{2}$.