If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel,then prove that the two lines are parallel.

(N/A) In the figure,a transversal $AD$ intersects two lines $PQ$ and $RS$ at points $B$ and $C$ respectively. Ray $BE$ is the bisector of $\angle ABQ$ and ray $CG$ is the bisector of $\angle BCS$; and $BE \parallel CG$.
We are to prove that $PQ \parallel RS$.
It is given that ray $BE$ is the bisector of $\angle ABQ$.
Therefore,$\angle ABE = \frac{1}{2} \angle ABQ$ ...... $(1)$
Similarly,ray $CG$ is the bisector of $\angle BCS$.
Therefore,$\angle BCG = \frac{1}{2} \angle BCS$ ...... $(2)$
But $BE \parallel CG$ and $AD$ is the transversal.
Therefore,$\angle ABE = \angle BCG$ (Corresponding angles axiom) ...... $(3)$
Substituting $(1)$ and $(2)$ in $(3)$,you get
$\frac{1}{2} \angle ABQ = \frac{1}{2} \angle BCS$
That is,$\angle ABQ = \angle BCS$
But,they are the corresponding angles formed by transversal $AD$ with $PQ$ and $RS$; and are equal.
Therefore,$PQ \parallel RS$ (Converse of corresponding angles axiom).