3. A.

As shown in Figure 3.
In $\triangle N B C$, applying Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{B M}{M N} \cdot \frac{N A}{A C} \cdot \frac{C D}{D B}=1 . \\
\text { Given } B D=D C \\
\Rightarrow \frac{B M}{M N}=\frac{A C}{A N} . \\
\text { Also, } \angle A B N=\angle A C B \text {, then } \\
\triangle A B C \backsim \triangle A N B \\
\Rightarrow \frac{A B}{A N}=\frac{A C}{A B}=\frac{B C}{B N} \Rightarrow \frac{A C}{A N}=\left(\frac{B C}{B N}\right)^{2} \\
\Rightarrow \frac{B M}{M N}=\frac{A C}{A N}=\left(\frac{B C}{B N}\right)^{2} \\
\Rightarrow \varphi=\lambda^{2} .
\end{array}
$$
3. Given that $M$ 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 a tangent to the circumcircle of $\triangle N B C$. Let $B C=\lambda B N, B M=\varphi M N$. Then the relationship between $\lambda$ and $\varphi$ is ( ).
(A) $\varphi=\lambda^{2}$
(B) $\varphi=\lambda^{2}-1$
(C) $\varphi=\frac{\lambda^{3}}{\lambda+1}$
(D) $\varphi=\lambda^{2}+1$