Fix a pointO. IfO is the midpoint of a line segment[XX′], then we say that X′ is a reflection of X across the pointO.
Note that the mapX7→X′is uniquely defined; it is called areflection acrossO. In this caseOis called thecenter of reflection. We assume that
O′ =O; that is, O is a reflection of itself across itself. If the reflection acrossOmoves a setS to itself, then we say thatSiscentrally symmetric with respect to O.
Recall that any motion is either direct or indirect; that is, it either preserves or reverts signs angles (see page 37).
7.6. Proposition.X Any reflection across a point is a direct motion. Proof. Observe that if X′ is a reflection of X across 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.
O
X
X′ Y
Y′ Fix two points X and Y; let X′ and Y′ be their reflections across O. To check that the re- flection is distance preserving, we need to show that X′Y′ =XY.
We may assume thatX,Y andOare distinct; otherwise the statement is trivial. By definition of reflection across O, we have thatOX =OX′,
OY =OY′, and the anglesXOY andX′OY′ are vertical; in particular ∡XOY = ∡X′OY′. By SAS,△XOY ∼=△X′OY′; thereforeX′Y′ =XY.
Finally, the reflection across O cannot be indirect since ∡XOY = =∡X′OY′; therefore it is a direct motion.
7.7. Exercise. Suppose∠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 5.7 to build another proof of Propo- sition 7.6. P Q O ℓ m
7.8. Theorem. Let ℓ be a line,Q∈ℓ, andP is arbi- trary point. SupposeO is the midpoint of[P Q]. Then a linempassing thruP is parallel toℓif and only if mis a reflection of ℓacross O.
Proof; “if ” part. Assumem is a reflection ofℓ acrossO. Suppose ℓ∦m; that isℓandmintersect at a single pointZ. Denote byZ′be the reflection ofZ acrossO.
Z Z′ O P Q m ℓ
Note that Z′ lies on both lines ℓ and m. It follows that Z′ = Z or equivalentlyZ =O. In this case O ∈ ℓ and therefore the reflection of ℓ across O is ℓ itself; that is, ℓ = m and in particular ℓ k m — a contradiction.
“Only-if ” part. Letℓ′ be the reflection ofℓ acrossO. According to the “if” part of the theorem,ℓ′ k ℓ. Note that both linesℓ′ andm pass thru
P. By uniqueness of parallel line (7.2), ifmkℓ, thenℓ′=m; whence the statement follows. A B C D
Transversal property
If the linetintersects each lineℓ and mat one point, then we say that t is a transversal to ℓ and m. For example, on the diagram, line(CB)is a transversal to (AB)and(CD).
7.9. Transversal property. (AB)k(CD) if and only if
➊ 2·(∡ABC+∡BCD)≡0.
Equivalently
∡ABC+∡BCD≡0 or ∡ABC+∡BCD≡π.
Moreover, if (AB) 6= (CD), then in the first case A and D lie on the opposite sides of(BC), in the second caseAandD lie on the same sides of(BC).
Proof; “only-if ” part. Denote byO the midpoint of[BC].
Assume(AB)k(CD). According to Theorem 7.8,(CD)is a reflection of(AB)acrossO. A A′ B C D O
Let A′ be the reflection of A across O. Then
A′∈(CD)and by Proposition 7.6 we have that
➋ ∡ABO=∡A′CO.
Note that
SinceA′,CandD lie on one line, Exercise 2.11 implis that
➍ 2·∡BCD≡2·∡BCA′.
Finally note that ➋, ➌, and➍imply➊.
“If ”-part. Given pointsA, B, andC there is unique line (CD)such that
➊holds. Indeed, suppose there are two such lines(CD)and(CD′), then 2·(∡ABC+∡BCD)≡2·(∡ABC+∡BCD′)≡0.
Therefore 2·∡BCD ≡ 2·∡BCD′ and by Exercise 2.11, D′ ∈ (CD), or equivalently the line(CD)coincides with the line(CD′).
From the “only-if” part we know that if ➊holds then(CD)k(AB).
By Theorem 7.2 there is unique line thru C that is parallel to (AB). Therefore it must be the same line (CD)such that➊hold.
Finally, if (AB) 6= (CD) and A and D lie on the opposite sides of (BC), then ∠ABC and∠BCD have opposite signs. Therefore
−π <∡ABC+∡BCD < π.
Applying➊, we get∡ABC+∡BCD= 0.
Similarly ifA andD lie on the same side of(BC), then ∠ABC and
∠BCD the same sign. Therefore
0<|∡ABC+∡BCD|<2·π
and➊implies that∡ABC+∡BCD≡π.
7.10. Exercise. Let △ABC be a nondegenerate triangle and P lies be- tweenAandB. Suppose that a lineℓpasses thruP and parallel to(AC). Show that ℓcrosses the side[BC] at another point, sayQ, and
△ABC ∼ △P BQ. In particular, P B AB = QB CB.
7.11. Exercise. Trisect a given segment with a ruler and a compass.