Each of the resistances in the network shown in fig 

in equal to ,R. The resistance between the terminals A and B is

The equivalent circuit is shown the fig.
 

Since the Wheatstone's bridge is balanced, therefore no current will flow through the arm KL. 

Equivalent resistance between AKM = R+R= 2R 

Equivalent resistance between ALM = R+R= 2R 

The two resistances are in parallel. 

Hence equivalent resistance between A and B is given by
$\begin{array}{l}\frac{1}{{\mathrm{R}}^{\mathrm{\prime }}}=\frac{1}{2\mathrm{R}}+\frac{1}{2\mathrm{R}}=\frac{2}{2\mathrm{R}}=\frac{1}{\mathrm{R}}\\ {\mathrm{R}}^{\mathrm{\prime }}=\mathrm{R}\end{array}$