In a voltage divider supplying a steady proportional voltage to a high resistance load, applying an AC signal with an AC coupling capacitor superimposes the voltages. Without the cap, only the signal gets through, and the DC divider does not at all. How does the cap achieve this?

 

No, that is not what happens. THe DC voltage on connecting the AC source will depend on the output "resistance" (impedence) of the device providing the AC signal.
Normally, all signal sources have a low output impedence. When the source is connected to the voltage divider point, it effectively shorts the DC to ground.
When you connect a capacitor, the DC does not get shorted to ground because the capacitor is "open" to DC. If you use a oscilloscope to monitor that junction, you will see a small dip in your DC as the capacitor charges to the divider voltage.

 

Thank you for the quick response! That explains everything.