Just to expand on that, in case you're not familiar, the principle is to produce a potential divider which drops the mains voltage to the required output voltage.
In a "normal" potential divider Vout = Vin * R2 / (R1+R2) where R2 is the resistor that is connected across the load - R1 and R2 appear in series across the supply.
In this design, R2 is substituted for the combined reactance of R39, R42, ZD6 and the load on the DC side of the bridge.... the clever bit is that R1 is substituted for a capacitor. Actually, in this case, it's substituted for the series combination of C32, C39, R36 and R47.
But why is a capacitor substituted for a resistor? Since it is an AC circuit with a known frequency, the capacitor has a known reactance, Xc=1/(2*pi*f*C) and so, as far as the "potential divider" action of the circuit is concerned, it behaves as a resistor of this value. Unlike a resistor, however, it does not dissipate any power. And that's why this circuit is often used - it is a simple way to drop voltage without dissipating an impractical amount of power.
Hope this helps!