In the given figure,$AX$ and $CY$ are respectively the bisectors of the opposite angles $A$ and $C$ of a parallelogram $ABCD$. Show that $AX \parallel CY$.

(N/A) Given: $ABCD$ is a parallelogram where $AX$ bisects $\angle A$ and $CY$ bisects $\angle C$.
$1$. Since $ABCD$ is a parallelogram,$\angle A = \angle C$ (opposite angles of a parallelogram are equal).
$2$. Dividing both sides by $2$,we get $\frac{1}{2} \angle A = \frac{1}{2} \angle C$.
$3$. Since $AX$ and $CY$ are bisectors,this implies $\angle XAB = \angle YCD$.
$4$. Also,in parallelogram $ABCD$,$AB \parallel DC$. Therefore,$\angle XAB = \angle AXC$ (alternate interior angles).
$5$. From steps $3$ and $4$,$\angle AXC = \angle YCD$.
$6$. Since these are corresponding angles for lines $AX$ and $CY$ with transversal $DC$,$AX \parallel CY$.