Problem — Bisectors of vertical angles

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Two lines meet at the point $O$, determining two pairs of vertical (opposite-at-vertex) angles. Prove that the bisectors of two vertical angles are opposite rays, i.e. they lie on a single straight line.

Solution

Let $OA$ and $OA'$ be two opposite rays (first line) and $OB$, $OB'$ two opposite rays (second line). The angles $\widehat{AOB}$ and $\widehat{A'OB'}$ are vertical, hence congruent; set $\widehat{AOB} = \gamma$.

Let $Os$ be the bisector of $\widehat{AOB}$ and $Os'$ the bisector of $\widehat{A'OB'}$. Then $\widehat{sOB} = \frac{\gamma}{2}, \qquad \widehat{B'Os'} = \frac{1}{2}\,\widehat{A'OB'} = \frac{\gamma}{2}.$

Consider the angle $\widehat{sOs'}$ obtained by passing from $Os$ to $Os'$ through the rays $OB$ and $OA'$: $\widehat{sOs'} = \widehat{sOB} + \widehat{BOA'} + \widehat{A'Os'}.$ Now $\widehat{BOA'}$ is the angle formed by $OB$ with the ray $OA'$, opposite to $OA$, hence the supplement of $\widehat{AOB}$: $\widehat{BOA'} = 180^\circ - \gamma.$ Moreover $\widehat{A'Os'} = \widehat{B'Os'} = \dfrac{\gamma}{2}$ (the bisector of $\widehat{A'OB'}$ is half an angle away from each of its sides). Adding up: $\widehat{sOs'} = \frac{\gamma}{2} + (180^\circ - \gamma) + \frac{\gamma}{2} = 180^\circ.$

The angle $\widehat{sOs'}$ is straight, hence $Os$ and $Os'$ are opposite rays and lie on the same line. $\blacksquare$