$A$ transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.

(N/A) Given: Two parallel lines $DE$ and $QR$ are intersected by a transversal at points $A$ and $B$ respectively. Let $AF$ be the bisector of $\angle CAE$ and $BP$ be the bisector of $\angle ABR$.
To prove: $AF \parallel BP$.
Proof: Since $DE \parallel QR$,the corresponding angles are equal,so $\angle CAE = \angle ABR$.
Multiplying both sides by $\frac{1}{2}$,we get $\frac{1}{2} \angle CAE = \frac{1}{2} \angle ABR$.
Since $AF$ and $BP$ are bisectors,$\frac{1}{2} \angle CAE = \angle FAB$ and $\frac{1}{2} \angle ABR = \angle ABP$.
Therefore,$\angle FAB = \angle ABP$.
These are corresponding angles formed by the transversal $n$ intersecting lines $AF$ and $BP$. Since the corresponding angles are equal,the lines must be parallel. Hence,$AF \parallel BP$.