(Almost) Wheatstone bridge
In theory, we have \$\text{R}_\text{measure}\to\infty\$, when that is the case we have the answer I wrote. You can use the same approach when that is not the case.

Well, we have the following circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
The input resistance can be found:
$$\text{R}_\text{in}=\frac{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\tag1$$
Now, the input current is given by:
$$\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_\text{in}}=\text{V}_\text{in}\cdot\frac{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag2$$
Now, the current \$\text{I}_1\$ is given by:
$$\text{I}_1=\frac{\text{R}_3+\text{R}_4}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\cdot\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_1+\text{R}_2}\tag3$$
Now, the current \$\text{I}_2\$ is given by:
$$\text{I}_2=\frac{\text{R}_1+\text{R}_2}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\cdot\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_3+\text{R}_4}\tag4$$
So, we get:

$$\text{V}_\text{A}=\text{I}_1\cdot\text{R}_2=\frac{\text{V}_\text{in}\text{R}_2}{\text{R}_1+\text{R}_2}\tag5$$
$$\text{V}_\text{B}=\text{I}_2\cdot\text{R}_4=\frac{\text{V}_\text{in}\text{R}_4}{\text{R}_3+\text{R}_4}\tag6$$

So, we also get:
$$\text{V}_\text{A}-\text{V}_\text{B}=\text{V}_\text{in}\cdot\left\{\frac{\text{R}_2}{\text{R}_1+\text{R}_2}-\frac{\text{R}_4}{\text{R}_3+\text{R}_4}\right\}\tag7$$