(N/A) First,draw the rhombus $ABCD$ in which $AB = 4 \, cm$ and $\angle ABC = 60^{\circ}$ as shown in the figure,and join $AC$.
Construct the triangle $AB'C'$ similar to $\triangle ABC$ with a scale factor of $\frac{2}{3}$ as instructed in the Mathematics Textbook for Class $X$.
Finally,draw the line segment $C'D'$ parallel to $CD$ such that $D'$ lies on $AD$.
Since $\triangle AB'C' \sim \triangle ABC$,we have $\frac{AB'}{AB} = \frac{AC'}{AC} = \frac{B'C'}{BC} = \frac{2}{3}$.
Since $C'D' \parallel CD$,$\triangle AD'C' \sim \triangle ADC$. Thus,$\frac{AD'}{AD} = \frac{AC'}{AC} = \frac{C'D'}{CD} = \frac{2}{3}$.
Since $ABCD$ is a rhombus,$AB = BC = CD = AD$.
Therefore,$AB' = B'C' = C'D' = AD' = \frac{2}{3} AB$.
Since all four sides of the quadrilateral $AB'C'D'$ are equal,$AB'C'D'$ is a rhombus.