Thevenin equivalent of voltage divider bias
I suspect that the load is the node looking into the base of the transistor. If I'm correct, then it boils down to changing the voltage divider (R1 and R2) into a single resistor in series with a new voltage source.

Firstly the new voltage source - it is the open circuit voltage at the junction of  R1 and R2 when the base is disconnected i.e. Vcc*(R2/(R1+R2)).

And the equivalent source resistance is R1 || R2 i.e. R1*R2/(R1+R2).

Anyway that's my take on it.