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# 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 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 $$O$$. 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 $$O$$. 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.