In the arrangement of resistances as shown, what is the equivalent resistance between A and B? 

We can simplify the circuit diagram

And we can see that between A and C terminal 2r and 2r resistances are parallel combination and we can calculate by using the formula

$\left(R_{eq}=\frac{R_1\times R_2}{R_1+R_2}\right)$

$\text{ Hence }\frac{2r\times 2r}{2r+2r}=r$

Similarly we can find the parallel combination between B and C terminal $\frac{2r\times 2r}{2r+2r}=r$

Again we can see that between C and D terminal r and r resistances are parallel combination and again we calculate by using formula $\left(R_{eq}=\frac{R_1\times R_2}{R_1+R_2}\right)$ hence we get $\frac{r\times r}{r+r}=\frac{r}{2}$

And now we get a simplifying circuit

Hence we can understand that it follows the wheatstone bridge concept and balance the following equation which is $\frac{P}{Q}=\frac{R}{S}$

where P = Q = R = S = r $\mathrm{\Omega }$ hence no current will be drawn from $\frac{r}{2}\mathrm{\Omega }$ resistance i.e.

Equivalent resistance is $\frac{2r\times 2r}{2r+2r}=r$

(Because r and r are in series combination so r + r = 2r)