How is the ac load line for the following amplifier derived?
This means the voltage drop across the equivalent resistance must be
  \$i_{c}R′_{L}\$


No, you forgot that at DC, the transformer primary has a DC resistance of circa 0 ohms and this is in parallel with \$R'_L\$. 

For AC signals of sufficient frequency that the primary inductance can be ignored (i.e. is much bigger in magnitude than \$R'_L\$), the AC signal at the collector is basically swinging symmetrically around \$+V_{CC}\$.

In other words a signal at the collector can swing from nearly 0 volts to +20 volts when \$+V_{CC}\$ is 10 volts: -



Picture taken from this slide show, page 16.