While your question is about the estimation of the variable resistor, and is answered by @Neil_UK I want to agree that this is a rather unpractical method for a precision measurement (and precision measurement is about the only reason you'd go the lengths of a Wheatstone bridge).
In reality, the Wheatstone bridge can be only really useful, if you have three known resistors, one of which R3 must be comparable to your unknown resistor R4 and the other two RM1 & RM2 should be closely matched.
simulate this circuit – Schematic created using CircuitLab
That allows you reduce the differential input voltage by orders of magnitude while the signal magnitude due to changes in R4 remains the same, thus "amplifying" your relative resistance change by the matching ratio of the bridge.
However, as Tony explained a Wheatstone bridge is tricky. You need 3 known resistors, which means they must be low drift which is expensive. Moreover, they also must have a low noise index, because they each add uncorrelated noise to your measurement.
So let me actually highlight the limited usefulness of the Wheatstone bridge in the 21st century. A more modern approach is the following:
simulate this circuit
It only requires R3 to be really precise. The noise of the current source is completely irrelevant due to the ratiometric nature of measurement. The impedance of the current source is also not critical, so you can realize it by a voltage source in series with a resistor. You obtain the same precision as with the Wheatstone bridge by using a single low noise and low drift resistor. C1 is there to remove the DC bias point and its value is not so critical. You also only need a single ended measurement (lower noise again). In the end, the input signal magnitude will be the same as that of the Wheatstone bridge (modulated by the switching frequency), but you obtain lower noise, lower drift, lower cost.