I know it's possible to replace the left side of this circuit with a thevenin equivalent, but on the left we're able to find the base voltage by the voltage divider Vcc * (R2)/(R2 + R1), which is the same as the thevenin voltage.
Doesn't this change the base voltage of the transistor since there's some base current that goes through the Rth?
The voltage divider equation doesn't give the right voltage between the base and the ground, because the base current loads the divider and causes some voltage drop. Thevenin's equivalent circuit gives an easy way to estimate that drop. It's exactly = base current * Rth.
Thevenin's equivalent circuit is a mathematically proven full replacement when seen by the rest of the circuit. That's why it gives the right results. Every circuit which obeys Ohms and Kirchoff's laws can be replaced with Thevenin's equivalent circuit. When seen from those 2 nodes for which the equivalent is calculated, there's no difference between the equivalent circuit and the original.
Thevenin's equivalent circuit consists one voltage source and one linear impedance.
ADD: The questioner wanted more in his comment. So, let's have two 10 kOhm resistors as the voltage divider in your first circuit. Let us have battery voltage =12V. Without any output current (=when the wire to the base of the transistor is disconnected) the voltage divider would give 6V output. The 50% drop is caused by the current through the resistors. If we increase the current by adding some load, for example we connet the base back, the voltage drop increases. Theres now less than 6V left.
Let's assume we take exactly 0,1mA to some load such as the base of the transistor. The voltge left is 6V-0,1mA * 5kOhm = 5,5V. The loading causes 0,5V drop. How I calculated it so easily? I used the Thevenin Equivalent, of course!
The phenomena "load current causes voltage drop" is more exactly said "The loading causes voltage drop due the internal resistance of the supplying circuit"
"But on the left, we're able to find the base voltage by the voltage divider Vcc * (R2)/(R2 + R1), which is the same as the Thevenin voltage" - This statement is absolutely incorrect.
simulate this circuit – Schematic created using CircuitLab
For example, let's consider the circuit above. According to voltage divider law, the voltage of NODE1 will be;
\begin{equation} V_1 = \frac{100}{100+100}10 = 5 \space volts\end{equation}
Now, if you connect a load at NODE1, the voltage of that terminal will not be 5 volts. It will change. For example, you have added something of 50-ohm resistance (i.e. RL). A significant amount of current will start to flow through that and the voltage of that terminal will also be changed. The new voltage will be;
\begin{equation} V_1 = \frac{100||50}{100+(100||50)}10 = 2.5 \space volts\end{equation}
The BJT is acting just as the load resistance RL (actually it's much more complicated than a linear resistor). As soon as the BJT is connected to the voltage divider network, the voltage of that point will drop. As a current will flow through the base. As you don't know the base current, you have no idea about the exact base voltage.
On the other hand, while measuring the Thevenin voltage, it is considered that the load is not connected to the system (i.e. the BJT is not connected) which cases zero load current. So the Thevenin voltage of that terminal will be 5 volts (as RL doesn't exist here).
I guess you have understood. Thevenin circuit simplifies a complicated network connected across the load; considering that; the load is not connected. You should study more on Thevenin equivalent system.
The Thevenin equivalent of the base pair of resistors gives you the unloaded Thevenin voltage and the unloaded Thevenin resistance. Of course, the BJT base loads the divider. So there is indeed a drop, which as you say, changes the actual base voltage of the BJT.
You already know all the details, really. You have set up the equivalent but just didn't follow through.
Let's apply KVL. Start with the Thevenin voltage and work your way around to the \$R_E=1\:\textrm{k}\Omega\$ resistor's grounded end.
$$V_{TH}-I_B\cdot R_{TH} - V_{BE} - I_E\cdot R_E = 0\:\textrm{V}$$
You should be able to follow the logic needed to write that out. Just walk through it and I'm sure you will get the same thing.
Now. You know that \$I_E=\left(\beta+1\right)\: I_B\$, so:
$$\begin{align*} V_{TH}-I_B\cdot R_{TH} - V_{BE} - \left(\beta+1\right)\: I_B\cdot R_E &= 0\:\textrm{V}\\\\ V_{TH} - V_{BE} &= I_B\cdot R_{TH}+ \left(\beta+1\right)\: I_B\cdot R_E\\\\ V_{TH} - V_{BE}&= I_B\cdot \left(R_{TH}+ \left(\beta+1\right) R_E\right)\\\\ I_B &= \frac{V_{TH} - V_{BE}}{R_{TH}+ \left(\beta+1\right) R_E} \end{align*}$$
Now you can compute \$I_B\$. It will be a little different depending upon the value of \$V_{BE}\$ and \$\beta\$ you use, of course. And these values definitely vary depending upon the BJT (and even among the same part number and batch.) But it should be reasonably close.
You can now use this estimate of \$I_B\$, multiplying it by the Thevenin resistance you computed earlier, to work out the voltage at the base as: \$V_B=V_{TH}-I_B\cdot R_{TH}\$. (Noting that we are discussing NPN.) You can see this fact from your own right-hand picture! Obviously, that must be the case.
That's really all there is.
Well... Not exactly. If you like math you can replace the value of \$V_{BE}\$ with the equation for it based upon the base current (which looks like the Shockley diode equation.) But then the solution requires the LambertW function, which you probably aren't familiar with. And it doesn't really help that much with practical designs anyway. Besides, there's still the effects of temperature; variations in the saturation currents over device and temperature; variations of \$\beta\$ over device, temperature, and collector current; basewidth modulation and the Early Effect; and... well, life is a lot less complicated and almost as good if you just don't worry yourself. (Unless you like math. ;)
