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Mathematics LibreTexts

5.3: Uniqueness of a perpendicular

  • Page ID
    23607
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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 \(\ell\).

    Proof

    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, \(m = (PP')\).