Problem — Bisectors of alternate interior angles are parallel

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Two parallel lines $r$ and $s$ are cut by a transversal $m$, which meets $r$ at the point $A$ and $s$ at the point $B$. Consider a pair of alternate interior angles: one with vertex $A$ and one with vertex $B$. Draw the bisector $p$ of the first and the bisector $q$ of the second. Prove that $p$ and $q$ are parallel.

Solution

Since $r \parallel s$ and $m$ is a transversal, the two alternate interior angles with vertices $A$ and $B$ are congruent; let $2\gamma$ denote their common measure.

The bisector $p$ splits the angle at $A$ into two angles of measure $\gamma$; likewise the bisector $q$ splits the angle at $B$ into two angles of measure $\gamma$.

Now regard the line $AB$ (that is, the transversal $m$ itself) as a transversal of the two lines $p$ and $q$. The angle that $p$ makes with $AB$, measured on the region between the two parallels, equals $\gamma$; the angle that $q$ makes with $AB$, on the region between them and on the opposite side of $m$, also equals $\gamma$. These two angles are therefore congruent alternate interior angles.

By the parallelism criterion (if two lines cut by a transversal form congruent alternate interior angles, then they are parallel), we conclude that $p \parallel q$. $\;\blacksquare$