Bipolar Transistor's Input Resistance
Or they actually just add the resistances without caring about the dependent source for some other reason?


The \$g_m\$ source produces current in the C-E branch, not in the B-E branch. So it doesn't contribute to determining the B-E equivalent resistance.

If you wanted the equivalent input resistance of a common-emitter amplifier with emitter degeneration (rather than just the transistor itself) ,



then you would certainly need to consider the effect of the \$g_m\$ source.

In comments you mentioned you are interested in a circuit theory answer. One thing to notice is that the transistor model you presented is the DC form of the hybrid pi model. It is designed so that you can write

$$\begin{bmatrix}v_b \\ i_c\end{bmatrix} =
\begin{bmatrix}r_\pi &amp; 0 \\
               g_m &amp; 1/r_o\end{bmatrix}
\begin{bmatrix}i_b \\ v_c\end{bmatrix}$$

(using lower case variables because these are small signal variations around the bias point)

That is, the model is designed so that \$V_b\$ only depends on \$I_b\$, while \$I_c\$ depends only on \$I_b\$ and \$V_c\$. More to the point, the \$r_\pi\$ element only affects the B-E port, and \$g_m\$ and \$r_o\$ only affect the C-E port.