Another way to look at it is to realize that this is the diode configuration of a transistor. Think back to your bipolar transistor theory. A diode can be made from a transistor by tying the base to the collector, and the voltage drop is Vbe, which makes the emitter voltage Vc-Vbe.
This circuit is a FET diode basically. The same principle applies and the voltage drop is Vth : the threshold voltage for Vgs.
You can do a simple analysis to show the real formula and see why what they show is an approximation.
Take the current equation for a FET, Id=k(Vgs-Vth)^2.
Then realize that once the gate is connected to the drain, Vgs=Vds, hence Id=k(Vds-Vth)^2.
Then solve for Vds and you get Vds=Vth + sqrt(Id/k)
So, really the actual voltage depends on the drain current, but the variation is small for reasonable currents. This is similar to a bipolar diode where we say the voltage drop is 0.7 V (assuming it's silicon). Really the actual voltage drop on a diode (whether bipolar or FET based) depends on current.