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AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig. 6.11). Show that AP \parallel BQ.

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Solution.

Given: In the figure, l \parallel m, AP and BQ are the Bisectors of \angle EAB and \angle ABH
To prove: AP \parallel BQ
Proof: Since
l \parallel m and t is a transversal therefore
\angle EAB = \angle ABH          (alternate interior angles)


\frac{1}{2} \angle EAB = \frac{1}{2}\angle ABH   (Divide both sides by 2)

\angle PAB = \angle ABQ (AP & BQ are the bisectors of \angle EAB & \angle ABH)
Now consider, two lines AP and BQ with transversal AB
\angle PAB and \angle ABQ are alternate interior angles and these are equal.
Hence, AP \parallel BQ

Hence proved

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