How to choose nodes for Thevenin equivalent circuit
Some Variations of BJT CE Stages
I'd like to start by throwing a bunch of slight variations on a theme. All of these below are similar BJT CE stages with similar purposes and applications. However, someday you'll want to be aware of some of the differences between them. For now, I just want to toss them out so that you won't be surprised by any of them.




simulate this circuit &ndash; Schematic created using CircuitLab



simulate this circuit
All of the above are single-ended and AC-coupled. Some exhibit highly variable gain and some don't, some are highly sensitive to ambient temperature and some aren't, some have much higher input impedance and some have much lower input impedance, some provide more degrees of design freedom and some less so. So there are various reasons for these.
I've neglected many of the incarnations that were found when germanium BJTs were more common (they have their own problems.) And I've removed some of the circuit incarnations that include modest isolation from the supply rails or where a real designer actually worked to properly shape the response of the amplifier over frequency. Trying to catalog every possible incarnation would be impossible here, anyway. But once you get a few basic variations on the theme down, they will all look similar to you and you'll immediately recognize &quot;why this, not that&quot; among them for any particular circumstance.
Also, many of the problems of some of the above can be repaired through something called global NFB (global negative feedback.) This is yet another concept you'll need in order to understand systems that include multiple stages chained together, for example. But that's for a later time, I think.
Let's move on.
Cononical BJT CE Amplifier Stage
The following left-side schematic is the standard BJT CE amplifier form. On the right is the Thenenized form of the exact same circuit.



simulate this circuit
An advantage arrives because a resistor divider pair allows the added freedom of constructing a custom Thevenin voltage without needing to create another power supply rail. It always comes with a cost -- the Thevenin source resistance. But that's also under control because you can change the two base-pair resistors to whatever you want. So it's pretty nice.
The result is the right-hand schematic above. Which has a designed Thevenin voltage supply and a Thevenin source resistance. The point is the same -- biasing the BJT. But you have one additional degree of freedom now -- the Thevenin voltage -- which you wouldn't have if you just tied a resistor to \$V_\text{CC}\$ (which isn't adjustable as it is the main power supply rail.)
What Nodes to Analyze?
In the above right-side schematic there are three unknown nodes. The base, the emitter, and the collector.
You can simplify your worries a lot, though.

Assuming the BJT is operating in active mode (the collector has a higher voltage than the base), which is always the case for a wel-designed CE amplifier stage, the collector &quot;looks like&quot; a current source (infinite output impedance) and can be ignored when analyzing the circuit.
The base and emitter voltages are separated by just one diode drop (roughly speaking) and therefore analyzing one of them analyzes the other one.

So you really only need to solve the base voltage (or the emitter voltage) to get the other voltage.
This means it's not all that hard. You can use KVL:
$$V_\text{TH}-I_B\cdot R_\text{TH}-V_\text{BE} - I_E\cdot R_\text{E} = 0\:\text{V}$$
Since \$I_E=\left(\beta+1\right)I_B\$ it follows that:
$$I_B=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)R_\text{E}}$$
And with that in hand it also then follows that:
$$\begin{align*}V_B&amp;=V_\text{TH}-I_B\cdot R_\text{TH}\\\\V_E&amp;=V_B-V_\text{BE}\end{align*}$$
Knowing \$V_E\$ allows you to now compute \$I_E=\frac{V_E}{R_\text{E}}\$ and \$I_C=\frac{\beta}{\beta+1}I_E\$. Knowing \$I_C\$ allows you to work out the voltage drop across \$R_\text{C}\$ and therefore now the collector voltage.
So that's really all there is to it. However, an assumption into all this is an a priori assumption about the value of \$V_\text{BE}\$. This value actually depends upon the value of \$I_C\$. So once you compute \$I_C\$ you can then refine the value of \$V_\text{BE}\$ and re-compute everything. Do that a few times (iterate) and you've gotten close enough.
There is a closed solution approach using the LambertW (product-log) function. But we can avoid messing with that, for now. Just iterate and you'll be good enough for most uses.
In case you want it, the Shockley equation for the active mode BJT is:
$$I_C=I_\text{SAT}\left(\exp\left[\frac{V_\text{BE}}{\eta\, V_T}\right]-1\right)$$
Or, for iterative purposes, just compute \$I_C\$ from above (not the Shockley equation immediately above, but I mean by applying \$\beta\$ to your calculated \$I_B\$ above) and then re-compute \$V_\text{BE}\$ from this:
$$V_\text{BE}=\eta\, V_T\,\ln \left(1+\frac{I_C}{I_\text{SAT}}\right)$$
Then go re-compute \$I_C\$ and repeat the process until satisfied.
The value of \$\eta\$ is usually just 1 for small-signal BJTs that are usually used in BJT CE amplifiers like this one. (It's the emission co-efficient.) The value of \$V_T\$ (the thermal voltage) is usually taken to be \$26\:\text{mV}\$ (near room temperature.) And the value of \$I_\text{SAT}\$ is something you'll need to look up for the BJT. But as a rough estimate you could use something like \$2\times 10^{-14}\:\text{A}\$ and probably be okay.