Solving transistor circuits
As Dave says, the diagram is almost certainly wrong. The bottom of the 40k emitter resistor should be connected to -20V. The explanation is still confusing on its own, so I'll try to unpack it.

First, they compute the DC emitter current. They do this by ignoring the base-emitter voltage drop, which gives ~0 V at the emitter. Then they get the current through Ohm's Law:

$$I_E = \frac {0\ \mathrm V - -20\ \mathrm V} {40\ \mathrm{k\Omega}} = 0.5\ \mathrm{mA}$$

Now they do the AC analysis using the T model. (I'm not sure why they use the T model; everyone else seems to like the hybrid-pi model more.) Their next step is to calculate \$r_e\$, the emitter's input resistance. This is done using the formula:

$$r_e = \frac {V_T} {I_E} \approx \frac {25\ \mathrm{mV}} {0.5\ \mathrm{mA}} = 50\ \Omega$$

It's much better to approximate \$V_T\$ as 25.9 mV -- it's not like it costs you anything. Regardless, the next step is to compute the base's input resistance. The example assumes (without saying it) that \$C_1\$, \$C_2\$, and \$C_3\$ are all big enough to act like short circuits at the signal frequency. Thus, the 40k resistor is totally bypassed, which is pretty common in CE amplifiers. This means the base's input resistance only depends on \$r_e\$:

$$r_b = \beta r_e = 50 \cdot 50\ \Omega = 2500\ \Omega$$

The total AC input resistance thus consists of \$r_e\$ and the 10k resistor in parallel, which equals 2000 ohms. The total AC load consists of the 20k collector bias resistor in parallel with the 20k load resistor, which equals 10k ohms.

Now for the voltage gain. Since \$R_S\$ is zero and \$r_b\$ is huge, all of \$v_s\$ is dropped across \$r_e\$. This means the emitter voltage equals \$v_s\$. The emitter current, then, is:

$$i_e = \frac {v_s} {r_e}$$

The collector current and emitter current are roughly equal. This means that a change in the emitter current shows up as an equal change in the collector current:

$$i_c \approx i_e = \frac {v_s} {r_e}$$

The collector voltage is the product of the collector current and the total load resistance (negative because the current flows from AC ground to the collector):

$$v_c = -i_c R_L$$

Now we can combine all this to get the AC voltage gain:

$$i_c \approx \frac {v_s} {r_e}$$
$$v_c = -i_c R_L \approx -\frac {v_s} {r_e} R_L$$
$$A_V = \frac {v_c} {v_s} = -\frac {R_L} {r_e} = -\frac {10\ \mathrm {k\Omega}} {50\ \Omega} = -200$$

But since \$r_e\$ depends on \$i_e\$, I don't think this is a great approximation. It works much better if you don't bypass the emitter resistor.

Phew! Almost done. Next, they compute the power gain, which is the product of the voltage gain and current gain. They assume the current gain equals beta (50), but as Dave pointed out this is not really true. Some of the input current goes to the 10k base bias resistor and some of the output current goes to the 20k collector bias resistor. 

Finally, they compute the output voltage. It looks like there's a typo here -- the voltage gain should be 200, not 20. This gives:

$$V_{out} = A_V V_{in} = 200 \cdot 2\ \mathrm {mV_{RMS}} = 400\ \mathrm {mV_{RMS}}$$

This looks like a pretty terrible textbook (or slide set?), so I'll give you the general procedure for all transistor circuits:


Find the DC bias voltages and currents.
Using the results of #1, compute the parameters of a small-signal model.
Use the small-signal model to determine the AC gain, input resistance, etc.