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 $\ell$.

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, $m = (PP')$.