One. (50 points) Given that point $M$ is a point on the median $A D$ of $\triangle A B C$, the line $B M$ intersects side $A C$ at point $N$, and $A B$ is the tangent to the circumcircle of $\triangle N B C$. Let $\frac{B C}{B N}=\lambda$, try to find $\frac{B M}{M N}$ (expressed in terms of $\lambda$).
Proof: In $\triangle B C N$, by Menelaus' theorem, we have
$$
\frac{B M}{M N} \cdot \frac{N A}{A C} \cdot \frac{C D}{D B}=1
$$

Since $B D=D C$, it follows that $\frac{B M}{M N}=\frac{A C}{A N}$.
Given $\angle A B N=\angle A C B$, we know $\triangle A B N \backsim \triangle A C B$, thus
$$
\frac{A B}{A N}=\frac{A C}{A B}=\frac{C B}{B N} \text {. }
$$

Therefore, $\frac{A B}{A N} \cdot \frac{A C}{A B}=\left(\frac{C B}{B N}\right)^{2}$, which means $\frac{A C}{A N}=\left(\frac{B C}{B N}\right)^{2}$.
Hence, $\frac{B M}{M N}=\left(\frac{B C}{B N}\right)^{2}$. Given $\frac{B C}{B N}=\lambda$, it follows that $\frac{B M}{M N}=\lambda^{2}$.