Different ways to calculate resistances 
The question asks about the resistance between B and X.
Now, one perspective is that point X is similar to point Y in terms of p.ds, therefore $R_{BY} = R_{BX} = R/2 $. And this is actually what the answer book says.
However, I thought of it this way, if $V_{AB}=P$, we have the same p.d across ZY. Since the combined resistance between B and Y is $R/2$, then p.d across BY is $P/2$. In addition, if the current through AB and ZY is I, that through CX will be its half. Ultimately, calculating $R_{BX}$ gives ${P/2}\over{I/2}$, i.e $R$. Why is this line of thought wrong?
Also, is there a way of calculating this using Kirchhoff's laws?
 A: Let $V_{ab}=25$ and $R=10$.
You would get $I=1$ because the total resistance is $25 (10+5+10)$. $5$ is because the $2$ resistors are in parallel. Thus the voltage drops around the circuit are $10,5,$and $10$ volts and each resistor in the parallel portion has $0.5$ amps.
A: If there were $3$ resistors of equal resistance connected in series to a battery of emf $V$ volts then the potential difference across each resistor would be $\frac{V}{3}$.
In your case, you have $3$ resistors ($2$ of resistance $R$ and $1$ resistor of equivalent resistance $\frac{R}{2}$). So the potential drop across each resistor won't be the same.
$Req = \frac{5R}{2} ;\space I = \frac{2V}{5R} $
The potential drop across AB, BY and YZ will be $\frac{2V}{5}$, $\frac{V}{5}$ and $\frac{2V}{5}$ respectively.
Let the potential drop across AB be $P$ which according to the calculation is equal to $\frac{2V}{5}$. 
Then the potential drop across CX or BY will be $\frac{V}{5}$ or $\frac{P}{2}$. As far as the results are concerned, your results are right.
It's correct to assume that in a circuit if 3 resistors ($R$, $R$ and $\frac{R}{2}$) connected in series and current through them is same then, according to ohm's law $V=I.Req$, potential drop across resistor of resistance $R$ will be $\frac{2V}{5}$ (or $I.R$ or $\frac{VR}{Req}$) and $\frac{R}{2}$ will be $\frac{V}{5}$ (or $I.\frac{R}{2}$ or $\frac{VR}{2Req}$). 
