High Impedance Input does not drop all voltage applied. Why not?
It does drop voltage, just as you expect.  It's more of a "leakage current" in the input transistors, but it can be modeled as a resistor. 

The thing is, you are connecting a (assumption, 50 ohm)low impedance to a hi-z input. Hi-z means high impedance.  The datasheet will tell you how much, but let's assume it is 1Megaohm.

So if you still have a voltage divider:  3.3000V divided into 50ohm &amp; 1Mohm. 

The result, is 3.2998V. Certainly dropped some voltage, but it's low enough that you can hardly measure it, let alone care about the actual voltage.