Applying Thevenin's theorem between 3 points
The images you gave are IMO a kind of roundabout way of the following:



simulate this circuit &ndash; Schematic created using CircuitLab
This circuit is equivalent to the one shown in your figures. And calculating the Thevenin-equivalent of the boxed part will get you your answers:
$$e_T = \frac{R_{B1}}{R_{B1}+R_{B2}}\cdot V_{CC}$$
$$R_T = R_{B1}//R_{B2}$$
\$e_T\$ is the output voltage of the boxed part if Q1 is not connected, or in that case \$v_B = e_T\$. After connecting Q1, the voltage \$v_B\$ will be partially pulled to GND via \$B\$ -&gt; Base-Emitter-diode -&gt; RE -&gt; GND. Hence,
$$e_T &gt; v_B &gt; v_M=0V$$