Due to the asymptote equations of the hyperbola being $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$, we can set $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=\lambda .(\lambda$ $=0$ or 1 )
(i. $\left.k^{2}\right) e^{2}-2 k m a^{2} x-a^{2} m^{2}-\lambda a^{2} b^{2}=0$. This equation clearly has real solutions.

By Vieta's formulas, we know $x_{1}+x_{2}=\frac{2 k m a^{2}}{b^{2}-a^{2} k^{2}}$, i.e., $\frac{x_{1}+x_{2}}{2}=\frac{k m a^{2}}{b^{2}-a^{2} k^{2}}$,
which means the x-coordinate of the midpoint of the two intersection points of the line $y=k x+m$ and the curve $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=\lambda(\lambda=0$ or 1 ) is the same (independent of $\lambda$).
Therefore, the midpoints of segments $A B$ and $C D$ coincide, i.e., $|A C|=|B D|$.