Q:

Why is the voltage gain \$=\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=-\frac{R_L}{R_E}\$

I learn some knowledge of  common emitter from this website: https://www.electronics-tutorials.ws/amplifier/amp_2.html
But i am confused about this formula : voltage gain  \$=\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=-\frac{R_L}{R_E}\$,here is the schematic of common emitter below 

First,if we know the output and the input wave,we can easily understand why is \$\frac{V_{out}}{V_{in}}=-\frac{R_L}{R_E}\$,not \$\frac{V_{out}}{V_{in}}=+\frac{R_L}{R_E}\$,but if we don't know common emitter acts like an inverter amplifier,how can we know \$\frac{V_{out}}{V_{in}}=-\frac{R_L}{R_E} ?\$,because the opposite voltage means the opposite current direction,but i can't understand why \$I_C\$ and \$I_E\$ are opposite,is it because the \$I_C\$ flows "into" bjt and  \$I_E\$ flows "out of" bjt,so they are opposite?
Next is why is \$=\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=\frac{\Delta V_L}{\Delta V_{RE}}\$,not \$\frac{\Delta V_L}{\Delta V_B}=\frac{\Delta V_L}{\Delta V_{R2}}\$,i think the reason of it is because \$I_B\$ is a input current of bjt,and \$V_{RE}=I_E\times R_E=(I_B+I_C)\times R_E\$ ,and  \$V_{R2}\$ has no relation with \$ I_B\$ ,so we can't say \$\frac{V_{out}}{V_{in}}=\frac{\Delta V_L}{\Delta V_B}=\frac{\Delta V_L}{\Delta V_{R2}}\$,however,i don't know the reason i think is correct or wrong,if wrong,can anyone tell me the reason of it??

A:

The minus sign in equation Vout/Vin = - R_load/R_emitter may appear counterintuitive. To understand its origin, remember that the network analysis here is performed for a linear circuit whereas the transistor circuitry is inherently non-linear. In order to be able to use Ohm's and Kirchhoff's laws, the circuit designer makes a number of idealizations, starting with an assumption that the voltages/currents in the circuit can be written as sums of DC voltages/currents plus (small additions of) AC voltages/currents. Or, alternatively, s/he examines small variations of voltages/currents near their "quiescent" values.
So, to fully understand the network analysis of transistor circuits, I would recommend you to learn from textbooks that use transistor models with current controlled current sources (CCCS)/voltage controlled current sources (VCCS) etc., the said models used to calculate biasing, gain, other parameters of small signal transistor amplifiers.
As for the common collector amplifier of the electronics-tutorials.ws tutorial, notice that for all practical purposes we can assume that C1 capacitance is sufficiently high, so that the AC component of V_base voltage is equal to Vin (Vin is an AC voltage source, I assume). In the Vin source's frequency region in which the C2 capacitance is sufficiently small to neglect current flowing thru a capacitor branch, an AC component of the current I_emitter is practically equal to an AC component of the V_emitter voltage divided by the R_emitter resistance. In the small signal approximation, V_base - V_emitter is constant (usually about 0.7V), therefore ΔV_base = ΔV_emitter, and ΔI_emitter = ΔV_base / R_emitter. Within a 1 percent error (beta = 100) I_collector = I_emitter; the AC components of emitter and collector currents are also equal (ΔI_collector = ΔI_emitter).
Vout is a voltage at the collector output of the transistor, as well as its AC component ΔV_load. Ohm's law gives us a voltage across the load resistor R_load measured relative to a reference point which is +Vcc pin of power supply (this means current flows out of R_load resistor); hence the minus sign in the equation
ΔV_load = - ΔI_collector * R_load. 
We arrive at equation ΔV_load/ΔV_base = - R_load/R_emitter.

