Q:

What is the voltage divider rule?

Can anyone explain what the voltage divider rule is? How has the author of this book used it for analyzing the Voltage Divider Bias circuit for transistors? 
And can anyone explain how the two resistors are parallel? And how has the author assumed the battery to be a short circuit (as shown in Figure 4.28)?
Thanks a lot. 
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From above link:

A:

We are looking at an AC amplifier. 
The DC part of the amplifier is only about setting up the Q-Point, or Working Point of the transistor.
Once that is determined, the actual AC workings can be looked at. 
I would propose you get hold of a complete book on basic electronics. The text you showed us here expects quite a bit of knowledge in the reader.
But lets see how this works out: 

Fig. 4.25: We describe the circuit. 
Fig. 4.27: The step showing the DC Part of the input side
Fig. 4.28: What remains if we look at AC -> VCC is replaced with a shortcut. This is a way to calculate the internal resistance of the network we are looking at. This is only a hypothetical step in a mathematical technique. 
"Applying the Voltage Divider Rule", meaning no other thing than "seeing how in 4.28 the Resistances are in parallel, we calculate the parallel resistance using the well known formula. But lets call it a rule". 
This all happens expecting the reader knows how to apply the Thevenin Theorem, which doubtlessly was explained earlier in the book. 
Fig. 4.29 Lets draw the resulting Network, showing the black box we are transforming. 
Eq 4.30 Calculating Ib using the Theremin replacement, this also happens in DC. 
Fig. 4.30 Two steps in this one: First, 4.29 is replaced by the appropriate Thevenin replacement. Secondly, it is applied to the base of the transistor. 
Equation 4.31 has very little to do with the whole path so far, it is simply the summation of the voltages on the output side. 

In fact, Msr. Thevenin was a French pioneer of electronics, read the Wikipedia article. He was not at all mentioned in my studies as an electrical engineer. 

A:

(1) It MAY be of assistance to you to note that equation 4.29 in the above cited text COULD be called "the voltage divider rule" as it refers to R1, R2 and Vcc. ie changing what it says just slightly without changing the meaning:

Vout = Vin x R2 / (R1 + R2) 

ie R1 & R2 form a voltage divider and the above equation defines a "rule" of the result.
BUT
(2) There IS NO "voltage divider rule" as such. 
Even if somebody uses that term there is still no such rule.
 BECAUSE the terminology is much too general.
 That's even more general than saying eg "The Ohm's law rule"
 where you at least have some guide. 
If you have a specific question you should explain it clearly in words and not use general terms or few words or the real requirement is liable to be missed.
Added:
Re question:

Can you tell me how this 'rule' is derived? I'm a beginner so I didn't understand how he got the relation Vr2= (R2)(Vcc)/(R1+R2)

(1) Short answer.
 Voltage across each resistor is proportional to current in it (Ohm's law).
As current in both resistors
 = battery current
 = the same
 THEN the voltages across each resistor are proportional to their resistance value.
THIS IS THE KEY FACTOR THAT MAKES THIS WORK
Vout = Vr2 = ib x R2
 Vcc = Vr1+Vr2 = ib x R1 + ib x R2 = ib x (R1 + R2) 
So Vout / Vcc
 = Vr2 / (Vr1 + VR2)
 = ib x R2 / (ib x (R1 + R2)     )
 Cancel ib's
 Vout/Vcc= R2/(R1 + R2)
 Multiply both sides by Vcc.
 Vout = Vcc x R2 / (R1 + R2)
 QED. 
(2) Longer answer.
You MUST know Ohms law.
If you don't know Ohms law and it's various re arrangements, stop reading this now, drop all lse and learn it. Wikipedia and Google know all about it N time over  
... time lapse ... or no time at all as the case may be ...
 So we know you know Ohm's law. 
So - one version of Ohm's law says, as you know

V = i x R

ie the voltage drop across a resistor is equal to the value of the resistor multiplied by the current flowing in it.
Now look at fig 4-29

Take this circuit in isolation.
 The current from the battery flows from B+ at the top left of R1, via R1, then via R2 and back to B- and the bottom left. 
 Look at the diagram and be SURE that you agree with the above.
 Now, lets call the battery current Ib.  
Call the current in R1 I_R1.
 It can be seen "by inspection that I_R1 = Ib.
Call the current in R2 I_R2.
 It can be seen "by inspection that I_R2 = Ib.
So I_R1 = IR2 = Ib.
 ie the current is the same in each resistor and out of and into the battery.
Now, the voltage across R1 = VR1 is, based on Ohm's law = I_R1 x R1.
 And, the voltage across R2 = VR2 is, based on Ohm's law = I_R2 x R2.
BUT I_R1 = Ib and IR2 = Ib.  
So  VR1 = I_R1 x R1 = Ib x R1
 And VR2 = I_R2 x R2 = Ib x R2   
The ratio of VR2 / VR1 = Ib x R2 / Ib x R1 = R2/R1
 ie the voltages across the two resistors are proportional to their resistance values. 
Look at the diagram.
 Vbattery = Vcc
 Vcc = the voltage across R1 + the Voltage across R2
 Vcc = VR1 + VR2
 Vcc = ib x R1  + ib x R2
 Vcc = ib (R1 + R2)  
So
To determine the ratio Vout / Vcc:
Vout / Vcc  = V_R2 / Vcc
 = ib x R2 / ib (R1 + R2)
 but the ib's cancel so
 Vout/ Vcc = R2 / (R1 + R2)
 and rearranging
Vout = Vcc x R2 / (R1 + R2)
So the voltage across R2 compared to battery voltage = Vr2= (R2)(Vcc)/(R1+R2) 

