Angle Bisectors are Locus of Points Equidistant from Lines
Theorem
Let $\LL_1$ and $\LL_2$ be straight lines in the plane.
The locus of points which are equidistant from $\LL_1$ and $\LL_2$ are the angle bisectors of $\LL_1$ and $\LL_2$.
Proof
Let $A'SA$ and $B'SB$ be the straight lines $\LL_1$ and $\LL_2$ respectively, intersecting at the point $S$.
Let $E$ denote the set of points equidistant from both $\LL_1$ and $\LL_2$.
Let $F$ denote the set of points on the angle bisectors of $\LL_1$ and $\LL_2$.
We are to show that $E = F$.
First we show that $F \subseteq E$.
Let $P \in F$.
Without loss of generality, let $P$ be on the angle bisector of $\angle ASB$.
Drop perpendiculars $PM$ from $P$ to $SA$ and $PN$ from $P$ to $SB$.
Because:
- $\angle PSM = \angle PSN$
- $\angle PMS = \angle PNS$
- $PS$ is common
we have that:
- $\triangle PSM = \triangle PSN$
and so:
- $PM = PN$
That is, $P$ is equidistant from both $\LL_1$ and $\LL_2$ and so:
- $P \in E$
The same argument is used mutatis mutandis to show that an arbitrary point $P'$ on the angle bisector of $\angle ASB'$ is also equidistant from both $\LL_1$ and $\LL_2$.
That is:
- $P' \in E$
Hence:
- $F \subseteq E$
$\Box$
We see immediately that $S \in E$.
As $S$ is on the intersection of $\LL_1$ and $\LL_2$, $S$ is also on their angle bisectors.
Hence $S \in F$.
Let $P \in E$ such that $P \ne S$.
We note that $P \notin \LL_1$ and $P \notin \LL_2$ as that would mean trivially that $P \notin E$.
Drop perpendiculars $PM$ from $P$ to $SA$ and $PN$ from $P$ to $SB$.
With reference to the diagram, note that:
- $PN = PM$
- $\angle PMS = \angle PNS = 90 \degrees$
- $PS$ is common
- $SN^2 + PN^2 = PS^2 = SM^2 + PM^2$
But as $PM = PN$ it follows that:
- $SM = SN$
Thus $\triangle PSM$ and $\triangle PSN$ are congruent.
This means:
- $\angle PSN = \angle PSM$
and so $P$ is on one of the angle bisectors of $\LL_1$ and $\LL_2$.
That is:
- $P \in F$
Hence by definition of subset:
- $E \subseteq F$
$\Box$
We have:
- $E \subseteq F$
and:
- $F \subseteq E$
and so by definition of set equality:
- $E = F$
Hence the result.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The incentre