Thevenin's theorem BJT derivation question
As I wrote in the comments, you should get into the mental practice of converting every single pair of resistors, those pairs spanning from one voltage source to another voltage source, into their Thevenin equivalent.  You just "turn the crank." You should get to the point where you do it almost without thinking about it.

Let's take your first diagram and make it worse. It's still the same circuit, except that now this circuit requires three newly added power supply rail voltages: \$-7.5\:\text{V}\$, \$+7.5\:\text{V}\$, and \$+30\:\text{V}\$. But other than that annoyance, it's the same circuit.







simulate this circuit &ndash; Schematic created using CircuitLab

See all those pairs in the circuit, itself, shown on the upper left area?

I've computed the Thevenin equivalents of each of them on the right upper side. From this, and some imagination, I think you can see that the circuits are the same.

Just get into the practice of it.

For complicated cases, such as my added collector and emitter pairs, you can look at the very bottom of the above schematic. The equation is simple:

$$\begin{align*}
R_\text{TH} &amp;= \frac{R_A\cdot R_B}{R_A+R_B}\\\\
V_\text{TH} &amp;= \frac{V_B\cdot R_A + V_A\cdot R_B}{R_A+R_B}
\end{align*}$$

Just memorize it.

You can see in the above schematic at the bottom section where I draw the locations of each of these voltages and resistors, along with the new equivalent. That's really all there is to it.