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Mathematics LibreTexts

5.2: Perpendicular Bisector

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    Assume \(M\) is the midpoint of the segment \([AB]\); that is, \(M \in (AB)\) and \(AM = MB\).

    The line \(\ell\) that passes thru \(M\) and perpendicular to \((AB)\), is called the perpendicular bisector to the segment \([AB]\).

    Theorem \(\PageIndex{1}\)

    Given distinct points \(A\) and \(B\), all points equidistant from \(A\) and \(B\) and no others lie on the perpendicular bisector to \([AB]\).


    截屏2021-02-03 下午4.38.06.png

    Let \(M\) be the midpoint of \([AB]\).

    Assume \(PA = PB\) and \(P \ne M\). According to SSS (Theorem 4.4.1), \(\triangle AMP \cong \triangle BMP\). Hence

    \(\measuredangle AMP = \pm \measuredangle BMP.\)

    Since \(A \ne B\), we have "-" in the above formula. Further,

    \[\begin{array} {rcl} {\pi} & = & {\measuredangle AMB \equiv} \\ {} & \equiv & {\measuredangle AMP + \measuredangle PMB \equiv} \\ {} & \equiv & {2 \cdot \measuredangle AMP.} \end{array}\]

    That is, \(\measuredangle AMP = \pm \dfrac{\pi}{2}\). Therefore, \(P\) lies on the perpendicular bisector.

    To prove the converse, suppose \(P\) is any point on the perpendicular bisector to \([AB]\) and \(P \ne M\). Then \(\measuredangle AMP = \pm \dfrac{\pi}{2}\), \(\measuredangle BMP = \pm \dfrac{\pi}{2}\) and \(AM = BM\). By SAS, \(\triangle AMP \cong \triangle BMP\); in particular, \(AP = BP\).

    Exercise \(\PageIndex{1}\)

    Let \(\ell\) be the perpendicular bisector to the segment \([AB]\) and \(X\) be an arbitrary point on the plane.

    Show that \(AX < BX\) if and only if \(X\) and \(A\) lie on the same side from \(\ell\).


    截屏2021-02-03 下午4.48.18.png

    Assume \(X\) and \(A\) lie on the same side of \(\ell\).

    Note that \(A\) and \(B\) lie on opposite side of \(\ell\). Therefore, by Corollary 3.4.1, \([AX]\) does not intersect \(\ell\) and \([BX]\) intersects \(\ell\); suppose that \(Y\) denotes the intersection point.

    Note that \(BX = AY + YX \ge AX\). Since \(X \not\in \ell\), by Theorem \(\PageIndex{1}\) we have \(BX \ne BA\). Therefore \(BX > AX\).

    This way we proved the "if" part. To prove the "only if" part, you need to switch \(A\) and \(B\) and repeat the above argument.

    Exercise \(\PageIndex{2}\)

    Let \(ABC\) be a nondegenerate triangle. Show that 

    \(AC > BC \Leftrightarrow |\measuredangle ABC| > |\measuredangle CAB|.\)


    Apply Exercise \(\PageIndex{1}\), Theorem 4.2.1 and Exercise 3.1.2.