Why the voltage is assumed zero Confusing
If an ideal diode is conducting, the voltage across it is 0V (by the definition of ideal that this book seems to use). That means that the diode can be viewed as a short circuit, and the ground node on the left diode's anode is &quot;shorted&quot; through the two ideal diodes to the node at the right diode's anode. Hence \$V = 0\ \mathrm V\$.
This actually also works if you assume the other common definition of an ideal diode, that a conducting ideal diode has exactly 0.7 (or 0.65, or 0.6, or whatever) volts across it, because you'll then have two identical voltage sources in series, but in opposite directions, so the overall voltage comes out to be zero.
With real diodes, the forward voltage depends (slightly) on the forward current, and the two diodes in this circuit have different currents through them, so the voltage at \$V\$ will come out to be a few millivolts, though small enough to ignore for many purposes.