8.5: Equidistant Property

Recall that distance from a line $\ell$ to a point $P$ is defined as the distance from $P$ to its foot point on $\ell$.

Proposition $\PageIndex{1}$

Assume $\triangle ABC$ is not degenerate. Then a point $X$ lies on the bisector of external bisector of $\angle ABC$ if and only if $X$ is equidistant from the lines $(AB)$ and $(BC)$.

Proof

We can assume that $X$ does not lie on the union of $(AB)$ and $(BC)$. Otherwise the distance to one of the lines vanish; in this case $X = B$ is the only point equidistant from the two lines.

Let $Y$ and $Z$ be the reflections of $X$ across $(AB)$ and $(BC)$ respectively. Note that

$Y \ne Z$.

Otherwise both lines $(AB)$ and $(BC)$ are perpendicular bisectors of $[XY]$. that is, $(AB) = (BC)$ which is impossible since $\triangle ABC$ is not degeneate. By Proposition 5.4.1,

$XB = YB = ZB$.

Note that $X$ is equidistant from $(AB)$ and $(BC)$ if and only if $XY = XZ$. Applying SSS and then SAS, we get that

$\begin{array} {rcl} {XY} & = & {XZ.} \\ {} & \Updownarrow & {} \\ {\triangle BXY} & \cong & {\triangle BXZ.} \\ {} & \Updownarrow & {} \\ {\measuredangle XBY} & \equiv & {\pm \measuredangle BXZ.} \end{array}$

Since $Y \ne Z$, we get that $\measuredangle XBY \ne \measuredangle BXZ$; therefore

$$\measuredangle XBY = -\measuredangle BXZ.$$

By Proposition 5.4.1, $A$ lies on the bisector of $\angle XBY$ and $B$ lies on the bisector of $\angle XBZ$; that is,

$2 \cdot \measuredangle XBA \equiv \measuredangle XBY$, $2 \cdot \measuredangle XBC \equiv \measuredangle XBZ.$

By 8.5.1,

$2 \cdot \measuredangle XBA \equiv -2 \cdot \measuredangle XBC.$

The last identity means either

$\measuredangle XBA + \measuredangle XBC \equiv 0$ or $\measuredangle XBA + \measuredangle XBC \equiv \pi,$

and hence the result.