Why should the \$\frac{e_5}{2}=rl_5(I_5)\$,not \$e_5=rl_5(I_5)\$ in the DC voltage drop?
I can't say for sure what's in the mind of whoever drew that diagram and wrote those equations. The diagram doesn't use the usual conventions of electrical engineering, and the equations don't use the usual symbols of electrical engineering, at least as I was taught them.

But, it's possible they are assuming that the return current for each load must return to the generator by a similarly branching path, and on a similarly sized wire, as the source current. 

Then the voltage seen by each load will be reduced not just by the voltage drop in the source line, but also by the voltage drop in the return path (which we're assuming is equal to the drop in the source line). 

So that the drop as estimated by the resistance of the supply wire is doubled in its effect on the voltage received by the load.

Edit

Here's how, if I'm guessing correctly about the meaning of your diagram, we'd represent a similar circuit with only 3 loads in a more standard schematic diagram:







simulate this circuit &ndash; Schematic created using CircuitLab

The main difference is I've included both the supply wires and the return wires.

The point is if you want to define the term \$e_1\$ so that the voltage across load 1 is given by \$V_1 = V_s - e_1\$ (where \$V_s\$ is the voltage at the source), then you need to account for resistive loss in both the supply wire and the return wire.