Voltage Divider Circuit Question
People's pedantic comments about voltage references and diode forward voltage are entertaining to read but fail to recognize the fundamental problem with your circuit: you're assuming that node "B" will be equal to 4.5V no matter what is connected to it — this assumption is wrong. Voltage dividers only provide a reference voltage that you can "look at" — the moment you start pulling current from a voltage divider, it will start to droop. To understand why, back up and remember that the voltage across a resistor is always proportional to the current going through it. So, if R1 and R2 are the same, and nothing else is attached to the circuit (i.e., the diode is gone):


the current flowing through R1 and R2 will be the same (since there's nowhere else for it to go),
thus the voltage across R1 and R2 will be the same (via Ohm's law), 
thus the voltage at node B will be 4.5V — half of the 9V supply.


But! The moment you introduce the LED to the circuit, some of R1's current will flow through R2, while the rest of R1's current will flow through the diode. Thus:


the current flowing through R1 and R2 will not be the same
thus the voltage across R1 and R2 will not be the same
thus the voltage at node B will not be 4.5V.


For practical circuits, voltage dividers are usually built with 10k-500k ohm resistors (to keep power consumption minimal). LEDs pull orders of magnitude more power than these resistors supply at "normal" operating voltages (such as your 9V supply). Thus, your voltage divider isn't going to just "droop a bit" — it will droop so far that I bet you won't measure more than a millivolt or so across the diode.

Using the Reference Voltage in Practical Circuits

Speaking generally, and ignoring the LED for now (which wouldn't typically be powered from a reference voltage), if you want to actually use the reference voltage you generate with a divider, you will need to feed that into some sort of buffer. This is the classic example of a unity-gain op-amp buffer circuit. Feed the voltage divider's output into the positive pin of the op-amp, and feed the output of the op-amp back into the negative terminal to provide feedback. Now, the op-amp's output will track the voltage divider perfectly — and you'll actually be able to source some current from the circuit to do something useful!

Powering LEDs

If your goal is simply to power an LED from the 9V battery, you'd typically use a series resistor (in R1's place) in the 330-1000 ohm range, and you'd have no need for a resistor in R2's spot. If this doesn't make sense to you, you should read up on nodal analysis, Kirchhoff's current and voltage laws, and Ohm's law.