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# 8.5: Equidistant Property

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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.