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5.3: Uniqueness of a perpendicular

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    Theorem \(\PageIndex{1}\)

    There is one and only one line that passes thru a given point \(P\) and is perpendicular to a given line \(\ell\).

    According to the above theorem, there is a unique point \(Q \in \ell\) such that \((QP) \perp \ell\). This point \(Q\) is called the foot point of \(P\) on \(\ell\).


    If \(P \in \ell\), then both existence and uniqueness follow from Axiom III.

    截屏2021-02-04 上午9.20.39.png

    Existence for \(P \not\in \ell\). Let \(A\) and \(B\) be two distinct points of \(\ell\). Choose \(P'\) so that \(AP' = AP\) and \(\meauredangle BAP' \equiv - \measuredangle BAP\). According to Axiom IV, \(\triangle AP'B \cong \triangle APB\). In particular, \(AP = AP'\) and \(BP = BP'\).

    According to Theorem 5.2.1, \(A\) and \(B\) lie on the perpendicular bisector to \([PP']\). In particular, \((PP') \perp (AB) = \ell\).

    Uniqueness for \(P \not\in \ell\). From above we can choose a point \(P'\) in such a way that \(\ell\) forms the perpendicular bisector to \([PP']\).

    截屏2021-02-04 上午9.28.00.png

    Assume \(m \perp \ell\) and \(P \in m\). Then \(m\) is a perpendicular bisector to some segment \([QQ']\) of \(\ell\); in particular, \(PQ = PQ'\).

    Since \(\ell\) is the perpendicular bisector to \([PP']\), we get that \(PQ = P'Q\) and \(PQ' = P'Q'\). Therefore,

    \(P'Q = PQ = PQ' = P'Q'\).

    By Theorem 5.2.1, \(P'\) lies on the perpendicular bisector to \([QQ']\), which is \(m\). By Axiom II, \(m = (PP')\).

    This page titled 5.3: Uniqueness of a perpendicular is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.