Question

If a traversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel , then P.T. the two lines are parallel.

Answer 1

In figure, transversal $AD$ intersects two lines $PQ$ and $RS$ at points $B$ and $C$ respectively.

Ray $BE$ is the bisector of $\angle ABQ$

So, $\angle ABE=\angle EBQ=\frac{1}{2}\left(\angle ABQ\right)$

and ray $CG$ is the bisector of $\angle BCS$

So, $\angle BCG=\angle GCS=\frac{1}{2}\left(\angle BCS\right)$

and $BE\parallel CG$

We should prove that $PQ\parallel RS$

Since, $BE\parallel CG$ and line $AD$ is the transversal,

$\angle ABE=\angle BCG$ ------ Corresponding angles

$\frac{1}{2}\left(\angle ABQ\right)=\frac{1}{2}\left(\angle BCS\right)$

$\angle ABQ=\angle BCS$

But, these angles are the corresponding angles formed by transversal $AD$ with $PQ$ and $RS$

We know that, If a transversal intersects two lines such that the pair of corresponding angles is equal, then lines are parallel to each other.

So, $PQ\parallel RS$

Hence Proved.