Problem — Bisectors of corresponding angles are parallel

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The parallel lines $u$ and $v$ are intersected by a transversal $n$, which meets them at the points $P$ and $Q$ respectively. Consider two corresponding angles, one with vertex $P$ and one with vertex $Q$, and draw their bisectors $g$ and $h$. Prove that $g$ and $h$ are parallel.

Solution

Since $u \parallel v$ and $n$ is a transversal, the two corresponding angles with vertices $P$ and $Q$ are congruent; let $2\delta$ denote their common measure.

The bisector $g$ splits the angle at $P$ into two equal parts, each of measure $\delta$; likewise the bisector $h$ splits the angle at $Q$ into two equal parts, each of measure $\delta$.

Take the line $PQ$ (that is, the transversal $n$) as a transversal of the two lines $g$ and $h$. Since the two original angles are corresponding and their bisectors occupy homologous positions with respect to $n$, the angle that $g$ makes with $n$ and the angle that $h$ makes with $n$, taken in the same position (both on the same side of $n$ and on the same side of the respective bisected lines), each measure $\delta$: they are therefore congruent corresponding angles.

By the parallelism criterion (if two lines cut by a transversal form congruent corresponding angles, then they are parallel), we obtain $g \parallel h$. $\;\blacksquare$