Voltage divider with short circuit part 2
But why is there a voltage on Vout?


By Ohm's law, the current through R2, from left to right, is

$$I_{R2} = \frac{V_A - V_{out}}{R_2}$$

Also, from Ohm's law

$$I_{R3} = \frac{V_{out} - V_B}{R_3} $$

Now, assuming there is no other circuit connected to the \$V_{out}\$ node, the current through R2 must be the current through R3, i.e., R2 and R3 are in series.  Thus

$$\frac{V_A - V_{out}}{R_2} = \frac{V_{out} - V_B}{R_3} $$

But, \$V_A = V_B = 2.5V\$ so the only possible solution is

$$V_{out} = 2.5V$$