7.2: Reflection Across a Point

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Fix a point $O$. If $O$ is the midpoint of a line segment $[XX']$, then we say that $X'$ is a reflection of $X$ across the point $O$.

Note that the map $X \mapsto X'$ is uniquely defined; it is called a reflection across $O$. In this case $O$ is called the center of reflection. We assume that $O' = O$; that is, $O$ is a reflection of itself across itself. If the reflection across $O$ moves a set $S$ to itself, then we say that $S$ is centrally symmetric with respect to $O$.

Recall that any motion is either direct or indirect; that is, it either preserves or reverts the signs of angles.

Proposition $\PageIndex{1}$

Any reflection across a point is a direct motion.

Proof

$截屏2021-02-09 上午9.44.46.png$

Observe that if $X'$ is a reflection of $X$ acroos $O$, then $X$ is a reflection of $X'$. In other words, the composition of the reflection with itself is the identity map. In particular, any reflection across a point is a bijection.

Fix two points $X$ and $Y$; let $X'$ and $Y'$ be their reflection across $O$. To check that the reflection is distance preserving, we need to show that $X'Y' = XY$.

We may assume that $X, Y$ and $O$ are distinct; otherwise the statement is trivial. By definition of reflection across $O$, we have that $OX = OX'$, $OY = OY'$, and the angles $XOY$ and $X'OY'$ are vertical; in particular $\measuredangle XOY = \measuredangle X'OY'$. By SAS, $\triangle XOY \cong \triangle X'OY'$; therefore $X'Y' = XY$.

Finally, the reflection across $O$ cannot be indirect since $\measuredangle XOY = \measuredangle X'OY'$; therefore it is a direct motion.

Exercise $\PageIndex{1}$

Suppose $\angle AOB$ is right. Show that the composition of reflections across the lines $(OA)$ and $(OB)$ is a reflection across $O$.

Use this statement and Corollary 5.4.1 to build another proof of Proposition $\PageIndex{1}$.

Theorem $\PageIndex{1}$

Let $\ell$ be a line, $Q \in \ell$, and $P$ is an arbitrary point. Suppose $O$ is the midpoint of $[PQ]$. Then a line $m$ passing thru $P$ is parallel to $\ell$ if and only if $m$ is a reflection of $\ell$ across $O$.

$截屏2021-02-09 上午9.52.01.png$

Proof

"if" part. Assume $m$ is a reflection of $\ell$ across $O$. Suppose $\ell \nparallel m$; that is $\ell$ and $m$ intersect at a single point $Z$. Denote by $Z'$ be the reflection of $Z$ across $O$.

$截屏2021-02-09 上午9.53.35.png$

Note that $Z'$ lies on both lines $\ell$ and $m$. It follows that $Z' = Z$ or equivalently $Z = O$. In this case $O \in \ell$ and therefore the reflection of $\ell$ across $O$ is $\ell$ itself; that is, $\ell = m$ and in particular $\ell \parallel m$ -- a contradiction.

"Only-if" part. Let $\ell '$ be the reflection of $\ell$ across $O$. According to the "if" part of the theorem, $\ell ' \parallel \ell$. Note that both lines $\ell '$ and $m$ pass thru $P$. By uniqueness of parallel lines (Theorem 7.1.1), if $m \parallel \ell$, then $\ell ' = m$; whence the statement follows.

Proposition \(\PageIndex{1}\)

Exercise \(\PageIndex{1}\)

Theorem \(\PageIndex{1}\)