Stiff Voltage Divider Bias for an Emitter Follower
First let me lead with a redrawn partial version of your circuit, which might be sufficient to convey the idea:







simulate this circuit &ndash; Schematic created using CircuitLab

Maybe that makes it easier to think about R1 and R2 in parallel. If not, read on...

Forget the transistor for now, and consider just a basic voltage divider, connected to some load:







simulate this circuit

Let's say that we want Vout = Vin/2, so we pick R1 and R2 to each be 1kΩ. What if RL is 250Ω?

R2 and RL in parallel are effectively:

$$ R_2 || R_L = \frac{1}{1/1\mathrm k\Omega + 1/250\Omega} = 200\Omega $$

Thus the actual behavior we get out of the voltage divider is:

$$ V_{out}
= \frac{R_2||R_L}{R_1 + R_2||R_L} V_{in}
= \frac{200\Omega}{1\mathrm k\Omega + 200\Omega} V_{in}
= \frac{V_{in}}{6} $$

This is not the \$V_{out} = V_{in}/2\$ that we wanted. It can be shown that what we actually got was equivalent to the voltage divider we wanted (\$V_{out} = V_{in}/2\$), in series with the Thévenin equivalent of the voltage divider (which is R1||R2), into the load:







simulate this circuit

Here we see that this is just another voltage divider, but without any load. See, we get the same result:

$$ V_{out} = \frac{V_{in}}{2} \frac{250\Omega}{250\Omega + 500\Omega}
= \frac{V_{in}}{2} \frac{1}{3} = \frac{V_{in}}{6} $$

Thus the rule of thumb for voltage dividers:

To make the error due to the load negligible, make the Thévenin equivalent resistance of the voltage divider at least 10 times smaller than the load.

When this rule is followed, then the current in the load is at least 10 times smaller than the current in the voltage divider, so the error introduced will be negligible.

Now, your transistor example is the same, but the current in RE is made smaller by a factor of \$h_{FE}\$. So, RL is equivalent to \$R_E \cdot h_{FE}\$. Otherwise we are just following the rule of thumb above about voltage dividers.

You can't ignore R1 because all the current through R2 or into the base must also go through R1. If you make either RE or R2 smaller, more current must flow through R1, thus there must be more voltage across R1, which might mess up your voltage divider if the change is large enough. The trick is to make the change in current due to variations of RE insignificant in comparison to the current already going through R1.

You could view R2 in parallel with your effective RL as the load for R1, and calculate your voltage divider based on that, but due to expected wide variations in \$h_{FE}\$, RL could in practice vary a lot. Thus, you want to design the circuit to be very insensitive to such variations.