Why is it possible to replace part of a BJT circuit with thevenin equivalent?
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"