Starting from the circuit schematic, as usually drawn here:
Something which is not shown is the decoupling capacitor from +12v to 0v.
The purpose of the decoupling capacitor is to remove any signals from the 12v line, in other words, it creates a very low impedance (short circuit) at signal frequencies.
The version of the circuit which you show, can be described as the "AC Equivalent Circuit", it is showing the circuit from the point of view of an AC signal.
The DC bias conditions are not shown.
So, consider R2 and Rc, each resistor has one end connected to the +12v line, but, the 12v line is short circuit to the 0v line as far as AC signals are concerned, and so that put R1 and R2 in parallel, and Rc is in parallel with RL (via C2).
So, consider R2 and Rc, each resistor has one end connected to the +12v line, but, the 12v line is short circuit to the 0v line as far as AC signals are concerned, and so that put R1 and R2 in parallel, and Rc is in parallel with RL (via C2).
It took me so long time to hear of something that intrigued me every time I run across input impedance of a BJT amplifier. Have to search and check with LTSpice. Gracias.
The article in the first post explains how easy it is to calculate the input impedance of a common-emitter transistor.
Simply calculate the base voltage to give the emitter voltage and emitter and collector current, then use the simple formula of (25/IE) times the beta..