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Unformatted text preview: Transformational Geomegy  Part II Theorems (NIB) 5.2 — 5.4 below fully characterize isometries of the Euclidean Plane.
They show that every isometry of the Euclidean plane is one of the following: (1) areﬂection,
(2) a composition of two reﬂections, or
(3) a composition of three reﬂections .
They also show that every isometry of the Euclidean plane is one of the following:
(1) a translation, (2) a rotation, (3) a reﬂection , or (4) a glide reﬂection. Theorems (NIB) 5.2 and 5.3 are proved in Neutral Geometry so that they may also be
used in the study of Hyperbolic Geometry. Theorem g NIB) 5.2 (The Congment Triangles Isomet_ry Theorem): Suppose that points A, B and C, and points D, E, and F are sets of three noncollinear
points each in Neutral Geometry such that AABC 2 ADEF. Then, there is an isometry t which maps AABC onto ADEF, that is, there is an isometry t such that t(A) = D, t(B) = E, and t(C) = F ,
and this transformation t can be chosen D = t(A)
such that at least one of the following is true: 3 t pl: _ ac)
1) t is a reﬂection about a mirror line, /\E= “B, 2) t is a composition of two reﬂections, 3) t is a composition of three reﬂections. c Proof of Theorem 5.2 The Con ent Trian les Isome Theorem : The proof is accomplished by constructing the isometry t as a composition of
reﬂections in a stepbystep manner, considering all cases of the possible placement of
the target triangle ADEF as it relates to the position of the source triangle AABC . The ﬁrst case to consider is the possibility that AABC and ADEF are names of the same
triangle. IfA=D andB=E andC=F, B E then we can choose t to be t = the identity transformation
where t(X) = X, for every point X inthe plane. Then, t is
an isometry such that
t(A)=A=D,t(B)=B=E, andt(C)=C=F.
In this case, t is the composition of two reﬂections, where both A = D
reﬂections have the same mirror line B . _
Forexample,ifwelet line e = 71; , then t = Re 0 Re. LI" 6 C=F Thus, we can assume that one of the points A, B, or C is distinct from D, E, or F,
respectively. Relabelling the points (if necessary), we assume that A 9e D .
Let line 8 be the perpendicular bisector of segment E . Let transformation t1 = R e . Since the mirror line e is the perpendicular bisector of 213 , t1 (A) = R e (A) = D , by
deﬁnition of reﬂections. Thus, A’ = t1(A) = D. LetB’= t1(B) and C’= t1(C). Now, t1 may or may not be the complete
transformation t we are seeking. It depends on the positions of E and F of
the target triangle ADEF . lfB'= E and C’= F, E=B'=t(3) then ADEF is positioned
o = A' = t(A) so that it is the mirror image AABC and
F = c' = t(C) we can choose t = t1,
Mirror lin e and we have proven the theorem A = th , perpendicular bisector ofzﬁ statement in the case that
B’ = E and C’ = F , where tis the reﬂection Re and
line ﬂ is the perpendicular bisector of segment 25 . Now, stillassumingthat t1 = Re, with A’ = t1(A), B' = t1(B) and C’= t1(C)
andwith A 75 D but with A’ = D, weknowthat B'= E or B'96 E. We assume ﬁrst that B ’ = E . In this case , we can assume that C ’ 7i F, because
the theorem statement was just proved to be true if C ’ = F . Thus,inthis case,weareassumingthat A’ = D , B’ = E, and C’ 75 F.
Let line 11 be the perpendicular bisector of the segment 5’? . Let t 2 = Rn .
Thus, t2( C') = RI1 (CI) = F , by the deﬁnition of reﬂection. We continue the construction of the isometry t by following the reﬂection t1 by the
reﬂection t2. Thus, we consider the transformation t2 0 t1. W6WlllWI‘lt6 A”: (tzo t1)(A) , B”: (tzo t1)(B) , CH: (tzo NOW, A”: t2(t1(A)) = t2(D) and B”: t2(t1(13)) = t203) Weshowthat t2(D) = D and that t2(E) = E byshowingthatD and E
are points on the mirror line n. 25 _ 5? since
AABC .2. ADEF , by assumption. I2 ll :15 ; A'C’ _ DEsince t1
isanisometry. D is equidistant from C l and
F , and so, by Theorem 3.2.8, D
is a point on line 11, the
perpendicular bisector of C'F . Mirror Line n =
the perpendiculaL
bisector of C'F « rpendicular__
' isector of AD Similarly, F5 2 5 since
AABC z ADEF , by assumption.
EE B'C’ _ E—C' since t1 is an isometry. III
II E is equidistant from C ’ and F , and so, by Theorem 3.2.8, B is a point on line 11, the
perpendicular bisector of C’F . Therefore, D and E are ﬁxed points of the reﬂection t2 = Rn since both points are
on the mirror line n. Therefore, t; (D) = D and t; (E) = E . Now, A”= (tzot1)(A) = t2(t1(A)) = t2(A’) = t2(D) = D.
E. Also, B”: (tzot1)(B) = t2(t1(B)) = t2(B’) = mm = Since the mirror line 11 is the perpendicular bisector of C'F, t2(C’) = Rn(C’) = F.
..c”= (t20t1)(C) = t2(t1(C)) = t2(C') = F.
Thus, wecanchoose t = (t2 0 t1). .'.t(A) = D, t(B) = E, and t(C) = F. Thus, we have proved the theorem
statement in the case that A’=D, B’= E and CliF and the isometry t is the
composition of two reﬂections: Re followed by Rn. Recall that we have assumed that A at D and, by deﬁnition of t1, A] = D. We have proved the theorem statement under these conditions in all cases for which
B’ = E. Thus, weassumethat A’ = D and B’ i E. Let line m be the perpendicular bisector of the segment E5 . Let t3 = RIn .
Thus, t3( B ’ ) = Rm (B ’) = E , by the deﬁnition of reﬂection. We continue the construction of the isometry t by following the reﬂection t1 by the
reﬂection t3. Thus, we consider the transformation t3 0 t1. Wewillwrite A”: (t30 t1)(A) , B”: (tgo t1)(B) , C”= (t3o t1)(C).
NOW, A”: t3(t1 = t3(D) and CH: t3(t1 = t3 We show that t3 (D ) = D and by showing that D is a point on the mirror line m. . B 5 DE since
AABC ; ADEF,
by assumption. AB .=. A’ ' E DB'
since t1 is an Perpendicular;
isometry. bisector of B'E . F D is equidistant
from B ’ and E , and so, by Theorem 3.2.8, C D is a point on line m, the perpendicular bisector of B71? .
Therefore, D is a ﬁxed point of the reﬂection t 3 = Rm since D is on the mirror line n.
Therefore, A" = t3 (A’) = t3 (D) = D and,
bydeﬂnitionof t3 , B" = t3 (B’) = E. Now, either C”= t3(C’) = F or C”= t3(C/) 75 F . If C” = t3 (C’) = F , then wecanchoose t = (t3 0 t1) becauseofthe
following calculations MA) = A” = (tsot1)(A) = t3(t1(A)) = t3(D) = D, t(B) = B” = (t30t1)(B) = t3(t1(B)) = ts(B') = E, t(C) = C” = (t30t1)(C) = t3(t1(C)) = t3(C’) = F,
provethat the theorem statement
is true in this case, and the isometry t
Perpendicular is the composition of “sector " ' two reﬂections :  dicular t = RmOR‘ ' A rofAB. This leaves one ﬁnal case to consider and that is the case in which
C” ¢ F , where C" = (t3 0 t1)(C) , which is illustrated in the
ﬁgure to the right : A]: = D, I t1(B) i E, / t1(C) 9 // = B
C
A” = t3(AI) = D,
B E ,
C N = t3(C/) i This is the situation for which the desired isometry t is a composition of 3 reﬂections. Let line k be the perpendicular bisector of the segment W . Let t4 = Rk .
Thus, t4( C H) = Rn (C H) = F , by the deﬁnition of reﬂection. We continue the construction of the isometry t by following the composition of
reﬂections t 3 o t 1 by the reﬂection t4. Thus, let isometry t be t= t4ot3ot1. B /$
t(C) = (t40t30t1XC)
= t4(t3(t1(c))) = t4(t3(C/)) t(C) = t4(C”) = F. e k = the perpendicular __
bisector of C"F . H
II t4° t3°t1
ltremainsonlytoshowthatt(A) = D and t(B) = E. Now,t(A) = t4(A”) = t4(D) and t(B) = t4(B”) = t4(E).
Once we have shown that D andE are ﬁxed points of t4, we can prove that t (A) = D
and t(B) = E. E a 5 since AABC ; ADEF, by assumption. Also, A ; A"C” 2 DC" since
t3 0 t1 is anisometry. D is equidistant from C” and F . D is apoint on line k, the perpendicular bisector of 57}: . by
Theorem 3.2.8 . E E E15 since AABC _=. ADEF, by
assumption. E5 ; B"C" EC"
since t 3 0 t1 is anisometry. Ill E is equidistant from C” and F. E is a point on line k, the perpendicular
bisector of C"F , by Theorem 3.2.8 . A Since D and E are on line k, the perpendicular bisector of 571? ,
t4(D) = Rk(D) = D and t4(E) = Rk(E) = E.
Thus,t(A) = t4(A”) = t4(D) = D and
t(B) = t4(B”) = t4(E) = E.
Recallthatt(C) = t4(C”) = F.
Thus, when t is the isometry
t = t40t30t1, o
t(A) = D, B W
t(B) = E,and
t(C) = E, and t maps AABC onto ADEF. Therefore,
the theorem statement
is true in this case
because t is the composition of 3 reﬂections: t RkORmORe. Therefore, the theorem is true in general because it has been proved to be true in all
possible cases. Q E D Theorem (MB) 5.3 (The Isometries from Reﬂections Theorem):
Given any isomtery t of the Neutral plane, then at least one of the following is true: 1) t is a reﬂection about a mirror line,
2) t is a composition of two reﬂections, 3) t is a composition of three reﬂections. Proof: Let t be any isometry of the Neutral plane.
Write X’ = t(X) for all points X.
Let A, B, and C be three non—collinear points.
Then, AABC 2 AA'B'C' by SSS, since t is an isometry.
By the Congruent Triangles Isometry Theorem (Theorem (NIB) 5.2),
there is an isometry to such that
to(A) = A' , t0(B) = B’, and t0(C) = C’,
and to is the composition of one, two, or three reﬂections in the plane.
Since t(A) = A’ = t0(A), t(B) = B’ = t0(B),
and t(C) = C’ = tom),
the isometries t and to agree at three noncollinear points in the plane.
Therefore, by the Three Points Agreement Theorem (Theorem 5.3.12) ,
as transformations, t = to .
Since to is the composition of one, two, or three reﬂections in the plane, t = to is also the composition of one, two, or three reﬂections in the plane . QED Theorem @132 5.4 (The Four Euclidean Isomeﬂ Types Theorem): Every isometry of the Euclidean Plane is one of the following types: 1) a translation TPQ for some vector PQ, 2) areﬂection Ra about amirror line e. 3) a rotation R 13,9 about a point P and through an angle 6 , or ————) 4) a glide reﬂection GL PQ’ e for some mirror line Z and some vector PQ
parallel to line (3 . Proof: (Incomplete) Let t be any isometry.
By the lsometries from Reﬂections Theorem (Theorem (NIB) 5.3),
at least one of the following is true:
1) t is a reﬂection about a mirror line,
2) t is a composition of two reﬂections, or
3) t is a composition of three reﬂections. As the work in a previous Lab showed, such reﬂections and their compositions result in
transformations of the plane of only four types: translations, reﬂections, rotations and glide reﬂections. The proof of this theorem is incomplete insofar as this last claim has only been
conjectured during that previous Lab and a proof of that claim is not provided here. QED (almost) 10 ...
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 Spring '10
 Shirley
 Euclidean geometry, perpendicular bisector, mirror line

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