Voltage drop assumption on diodes (LEDs, etc.)
Diodes have very steep output characteristics - there are almost no current below some voltage value, and after this voltage is reached the current increase is very steep:



This steep increase in current at \$V_d\$ essentially means that no matter what current you'll force through the diode, the voltage will not vary much from \$V_d\$. 

You can argue that for very large currents the voltage can increase by a significant amount, but in practice this will not happen. What will happen is that when the current increases over a nominal diode's current, the diode will be permanently damaged (short-circuit, open-circuit or some other bad effect).

Due to the above reasons, if the diode is present and is conducting - it will have a voltage drop of magnitude \$\approx V_d\$ across it.

Now, you may ask: "But how do we know a priori that the diode is conducting?". Good question. We really can't know this a priori, but we can ASSUME that this is the case and perform an analysis of the circuit. After the analysis is done we must ensure that all the voltages and the currents in our circuit are consistent. If they are - our assumption was correct. If, on the other hand, we find that the voltage drop (as follows from the analysis) on the diode is below its  \$V_d\$, then it is a contradiction and our assumption was wrong - the diode is in cut-off region of operation and behaves as an open circuit.