Proving the output voltage of a resistor ladder only with usage of currents
Redraw the schematic in a slightly less confusing way:







simulate this circuit &ndash; Schematic created using CircuitLab

Start at the bottom:

R5 and R6 have the same voltage across them, and have the same resistance so will share the current equally  (or if you prefer, they behave like a single resistor of resistance R). Calling the sum of the currents through R5 and R6, I, we have

V1 = I × R

Next consider R4: The current flowing out of the bottom of R4, must be the same as the current flowing into R5 and R6, which is just I.

V2 = V1 + I × R

Next consider R2 and (R1+R3). That same current, I, must be split between them. You know the voltage across R2 (of resistance 2R) is the same as the sum of the voltages  across R1 and R3 (of combined resistance 3R) 

using ohms law you know 2R × I_R2 = V_R2 = V_R1 + V_R3 = 2R × I_R1 + R × I_R3 = 3R × I_R3  (since the current through R1 and R3 is the same)

2 I_R2 = 3 I_R3

You also know that I_R2 + I_R3 = I (since the current flowing out the bottom of R3 and R2, can only flow into R4)

Solving those two equations you get

I_R2 = 3/5 I

I_R3 = 2/5 I

So you now know the voltage cross R2 is now 2R × 2/5 I.

which gives

V0 = V2 + 2R × 3/5 I = V1 + I × R + 2R × 3/5 I = I × R + I × R + 2R × 3/5 I = (2+6/5) × I × R = (16/5) × I × R

and 

Vout = V2 + R × 2/5 I = V1 + I × R + R × 2/5 I = I × R + I × R + R × 2/5 I = (12/5) x I × R

combining these gives

Vout = (12/5)/(16/5) V0 = (3/4) V0