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 $E = F$.


 * Bisectors-of-angles.png

First we show that $F \subseteq E$.

Let $P \in F$.

, 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 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:
 * $F \subseteq E$

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

By Pythagoras's Theorem:
 * $SN^2 + PN^2 = PS^2 = SM^2 + PM^2$

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:
 * $E \subseteq F$

We have:
 * $E \subseteq F$

and:
 * $F \subseteq E$

and so by definition of set equality:
 * $E = F$

Hence the result.