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 \(\ell\).
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
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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.