Draw a parallelogram $ABCD$ in which $BC = 5 \, cm$,$AB = 3 \, cm$ and $\angle ABC = 60^{\circ}$. Divide it into triangles $BCD$ and $ABD$ by the diagonal $BD$. Construct the triangle $BD'C'$ similar to $\triangle BDC$ with a scale factor of $\frac{4}{3}$. Draw the line segment $D'A'$ parallel to $DA$,where $A'$ lies on the extended side $BA$. Is $A'BCD'$ a parallelogram?

(A) $1.$ Draw a line segment $AB = 3 \, cm$.
$2.$ Draw a ray $BY$ making an angle $\angle ABY = 60^{\circ}$.
$3.$ With $B$ as the center and a radius of $5 \, cm$,draw an arc to cut the ray $BY$ at point $C$.
$4.$ Draw a ray $AZ$ parallel to $BY$ such that $\angle BAZ = 120^{\circ}$ (since $ABCD$ is a parallelogram,$\angle ABC + \angle BCD = 180^{\circ}$).
$5.$ With $A$ as the center and a radius of $5 \, cm$,draw an arc to cut the ray $AZ$ at point $D$.
$6.$ Join $CD$ to complete the parallelogram $ABCD$.
$7.$ Join $BD$,which is a diagonal of the parallelogram $ABCD$.
$8.$ From $B$,draw any ray $BX$ downwards making an acute angle $\angle CBX$.
$9.$ Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$10.$ Join $B_3C$ and draw a line through $B_4$ parallel to $B_3C$ intersecting the extended line segment $BC$ at $C'$.
$11.$ From point $C'$,draw $C'D' \parallel CD$ intersecting the extended line segment $BD$ at $D'$.
$12.$ Draw a line segment $D'A'$ parallel to $DA$,where $A'$ lies on the extended side $BA$.
$13.$ Yes,$A'BCD'$ is a parallelogram because by construction,$A'D' \parallel BC$ and $A'B \parallel D'C'$.