Download Chapter 2: Congruence

Chapter 2
Congruence
In this chapter we use isometries to study congruence. We shall prove the
Fundamental Theorem of Transformational Plane Geometry, which states that
two plane figures are congruent if and only if the first is the image of the second
under a composition of three or fewer reflections. This is a truly remarkable
fact. Our approach will be to study the fixed point sets for various families of
isometries. We shall observe that an isometry is completely determined by its
set of fixed points. This powerful idea enables us to identify all plane isometries
and understand their properties.
2.1
The Three Points Theorem
The facts that appear in our next three theorems are quite surprising:
Theorem 73 An isometry that fixes two distinct points P and Q, fixes every
←→
point on the line P Q.
Proof. Let P and Q be distinct fixed points of an isometry α and let R be
←→
any point on line P Q distinct from P and Q. Note that PR ∩ QR = R since P,
Q and R are distinct and collinear. But R0 = α (R) ∈ PR since P R = P R0 and
R0 = α (R) ∈ QR since QR = QR0 . Therefore R0 = R (see Figure 2.1).
39
40
CHAPTER 2. CONGRUENCE
Figure 2.1.
Theorem 74 An isometry that fixes three non-collinear points is the identity.
Proof. Let P, Q, and R be three non-collinear points that are fixed under
an isometry α. By Theorem 73, α fixes every point on 4P QR. Let Z be any
point in the plane off of 4P QR and choose a point M 6= Z interior to 4P QR
(see Figure 2.2).
Q
P
M
Z
R
Figure 2.2.
←−→
Then line ZM intersects 4P QR in two distinct points, which are fixed by α, so
←−→
each point on ZM is fixed by α, by Theorem 73. In particular, the arbitrarily
chosen point Z is fixed by α. Therefore α = ι, as claimed.
Theorem 75 (The Three Points Theorem) Two isometries that agree on
three non-collinear points are equal.
2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS
41
Proof. Suppose that α and β are isometries and P, Q, and R are noncollinear points such that
α(P ) = β(P ),
α(Q) = β(Q),
and α(R) = β(R).
(2.1)
Apply α−1 to both sides of each equation in (2.1) and obtain
P = (α−1 ◦ β)(P ),
Q = (α−1 ◦ β)(Q),
and R = (α−1 ◦ β)(R).
Thus α−1 ◦β is an isometry that fixes three non-collinear points and by Theorem
74, α−1 ◦ β = ι. Apply α to both sides of this last equation gives the desired
result.
In the next two sections, we use the Three Points Theorem to characterize
translations and rotations in a new and very important way.
2.2
Translations as Products of Reflections
Theorem 76 Let and m be distinct parallel lines and let n be a common
perpendicular. Let L = ∩ n and M = m ∩ n. Then σ m ◦ σ is the translation
by vector 2LM, i.e.,
σ m ◦ σ = τ 2LM .
Proof. Let L0 = σ m (L); then
(σ m ◦ σ )(L) = σ m (σ (L)) = σ m (L) = L0 = τ LL0 (L).
Let K be a point on
(2.2)
distinct from L, and let K 0 = τ LL0 (K). Then
τ KK0 = τ LL0
by Theorem 29. Therefore
(σ m ◦ σ )(K) = σm (σ (K)) = σ m (K) = K 0 = τ KK0 (K) = τ LL0 (K).
(2.3)
Let J = σ (M ); then L is the midpoint of J and M and M is the midpoint of L
and L0 . Hence JL = LM = ML0 so that JM = JL + LM = LM + ML0 = LL0
and
τ JM = τ LL0 .
Therefore
(σ m ◦ σ )(J) = σ m (σ (J)) = σ m (M ) = M = τ JM (J) = τ LL0 (J)
(see Figure 2.3).
(2.4)
42
CHAPTER 2. CONGRUENCE
Figure 2.3.
By equations (2.2), (2.3), and (2.4), the isometries σ m ◦ σ and τ LL0 agree on
three non-collinear points J, K, and L, so that
σ m ◦ σ = τ LL0
by Theorem 75. Since M is the midpoint of L and L0 we have
τ LL0 = τ 2LM ,
as desired.
In the proof above, M is the midpoint of LL0 so by Theorem 56 we have
τ LL0 = ϕM ◦ ϕL , which proves the next corollary:
Corollary 77 Let and m be parallel lines and let n be a common perpendicular. Let L = ∩ n and M = m ∩ n. Then
σm ◦ σ = ϕM ◦ ϕL .
Figure 2.4.
The next theorem tells us that the converse is also true:
2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS
43
Theorem 78 An isometry α is a translation if and only if α is a product of
two reflections in parallel lines.
Proof. Let be any line. Then α = ι (the identity transformation) if and
only if α = σ 2 , and the statement holds in this case. So assume that α 6= ι
and note that implication ⇐ was proved in Theorem 76. For the converse, let
L and N be distinct points and consider the non-identity translation τ LN ; we
must show that τ LN is a product of two reflections in distinct parallel lines. Let
←→
M be the midpoint of L and N ; let and m be the perpendiculars to LM at L
and M , respectively (see Figure 2.5).
P ' = σl (P)
P
P '' = σm (P ')
L
M
l
N
m
Figure 2.5.
Then by Theorem 56,
τ LN = ϕM ◦ ϕL
and by Corollary 77,
ϕM ◦ ϕL = σ m ◦ σ .
Therefore
τ LN = σ m ◦ σ
with km as desired.
Here is a useful trick that transforms a product of reflections in parallel lines
into a product of halfturns. Given parallel lines and m, choose any common
perpendicular n; let L = ∩ n and M = m ∩ n. Then
σ m ◦ σ = σ m ◦ ι ◦ σ = σ m ◦ (σ n ◦ σ n ) ◦ σ = (σ m ◦ σn ) ◦ (σ n ◦ σ ) = ϕM ◦ ϕL .
←→
Conversely, given distinct points L and M, let n = LM , let be the line through
L perpendicular to n, and let m be the line through M perpendicular to n. Then
reading the calculation above from right to left we see how to get from a product
of halfturns to a product of reflections in parallel lines.
If P, Q, R and S are points such that PQ = RS, then τ PQ = τ RS . Since
the product of two reflections in distinct parallel lines is a translation, there
44
CHAPTER 2. CONGRUENCE
is the following analogous statement for parallels: If lines p, q, r and s are
parallel lines passing through collinear points P, Q, R and S, respectively, then
σq ◦ σp = σs ◦ σr .
Theorem 79 Let , m and n be distinct parallel lines. There exist unique lines
p and q parallel to such that
σm ◦ σ = σn ◦ σp = σq ◦ σn .
Proof. Given distinct parallels , m and n, choose any common perpendicular b. By Theorem 58, there exist unique points P and Q on b such that
ϕP = ϕN ◦ ϕM ◦ ϕL and ϕQ = ϕM ◦ ϕL ◦ ϕN . Thus
ϕM ◦ ϕL = ϕN ◦ ϕP = ϕQ ◦ ϕN .
Let p and q be the lines perpendicular to b at P and Q, respectively (see Figure
2.6).
Figure 2.6.
Then by Corollary 77,
σ m ◦ σ = ϕM ◦ ϕL = ϕN ◦ ϕP = σ n ◦ σ p
and
σ m ◦ σ = ϕM ◦ ϕL = ϕQ ◦ ϕN = σ q ◦ σ n .
Thus given distinct parallels , m and n, Theorem 79 tells us that n determines unique parallels p and q such that σm ◦ σ = σ n ◦ σp = σ q ◦ σ n .
Furthermore, line p is the unique line parallel to n such that the directed distance from p to n equals the directed distance from to m; line q is the unique
line parallel to n such that the directed distance from n to q equals the directed
distance from to m. Thus:
2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS
45
Corollary 80 Let P and Q be distinct points and let be any line perpendicular
←→
to P Q. Then there exists a unique line m parallel to such that
τ PQ = σ m ◦ σ .
(See Figure 2.7.)
Q
P
M
L
m
l
Figure 2.7.
Corollary 81 Let , m, and n be distinct parallel lines. There exists a unique
line p parallel to , m, and n such that
σp = σn ◦ σm ◦ σ ,
i.e., the product of three reflections in parallel lines is a reflection in some unique
line parallel to them.
Proof. By Theorem 79, there exists a unique line p parallel to , m, and n
such that
σm ◦ σ = σn ◦ σp
(see Figure 2.8).
l
m
p
Figure 2.8.
n
46
CHAPTER 2. CONGRUENCE
Apply σ n to both sides and obtain
σp = σn ◦ σm ◦ σ .
Exercises
1. Lines and m have respective equations Y = 3 and Y = 5. Find the
equations of the translation σ m ◦ σ .
2. Lines and m have respective equations Y = X and Y = X + 4. Find the
equations of the translation σ m ◦ σ .
3. The vector of the translation τ is
m such that τ = σ m ◦ σ .
£
4
−3
¤
. Find the equations of lines
and
4. The translation τ has equations x0 = x + 6 and y 0 = y − 3. Find equations
of lines and m such that τ = σ m ◦ σ .
5. Lines , m and n have respective equations Y = 3, Y = 5 and Y = 9.
a. Find the equation of line p such that σ m ◦ σ = σ p ◦ σ n .
b. Find the equation of line q such that σ m ◦ σ = σ n ◦ σ q .
2.3
Rotations as Products of reflections
In the previous section we observed that every non-identity translation is a
product of two reflections in distinct parallel lines. In this section we prove the
analogous statement for rotations: Every non-identity rotation is the product
of two reflections in distinct intersecting lines. Thus translations and rotations
are remarkably similar and have many analogous properties.
Note that if and m are distinct intersecting lines, there is one directed
angle from and m with measure Θ1 in the range 0 < Θ1 < 180 and another
with measure Θ2 in the range 180 < Θ2 < 360. In fact, Θ2 = Θ1 + 180.
Theorem 82 Let and m be distinct lines intersecting at a point C and let
0◦ < Θ◦ < 180◦ be the measure of the angle from to m. Then
σ m ◦ σ = ρC,2Θ ,
for every Θ ∈ Θ◦ , i.e., σ m ◦ σ is the rotation about C through twice the directed
angle from to m.
2.3. ROTATIONS AS PRODUCTS OF REFLECTIONS
47
Proof. First observe that
(σ m ◦ σ ) (C) = σ m (σ (C)) = σ m (C) = C = ρC,2Θ (C).
(2.5)
◦
Let Θ ∈ Θ , let L be a point on distinct from C and let M be the point in
−→
−−→
m ∩ CL such that the directed angle measure from CL to CM is Θ. Let
L0 = ρC,2Θ (L);
then L0 lies on CL and m is the perpendicular bisector of LL0 (see Figure 2.9).
Therefore L0 = σ m (L) and we have
(σm ◦ σ ) (L) = σ m (σ (L)) = σ m (L) = L0 = ρC,2Θ (L).
(2.6)
Let J = σ (M ); then is the perpendicular bisector of JM , in which case J lies
−→
−−→
on CL and the directed angle measure from CJ to CM is 2Θ.
L ' = ρC,2Θ (L)
CL
m
M
Θ
C
Θ
Θ
L
l
J = σl (M)
Figure 2.9.
Hence
M = ρC,2Θ (J)
so that
(σ m ◦ σ ) (J) = σ m (σ (J)) = σ m (M ) = M = ρC,2Θ (J).
(2.7)
By equations (2.5), (2.6), and (2.7), the two isometries σ m ◦ σ and ρC,2Θ agree
on non-collinear points C, J and L. Therefore
σ m ◦ σ = ρC,2Θ
by Theorem 75, as claimed.
As with translations, the converse is also true:
48
CHAPTER 2. CONGRUENCE
Theorem 83 A non-identity isometry α is a rotation if and only if α is the
product of two reflections in distinct intersecting lines.
Proof. The implication ⇐ was proved in Theorem 82. For the converse,
given ρC,2Θ , let be any line through C and let m be the unique line through
C such that the directed angle measure from to m is Θ (see Figure 2.10).
Figure 2.10.
By Theorem 82, ρC,2Θ = σ m ◦ σ .
There is the following analogue to Theorem 79:
Theorem 84 Let , m, and n be distinct lines concurrent at point C. There
exist unique lines p and q concurrent at C such that
σm ◦ σ = σn ◦ σp = σq ◦ σn .
Proof. Given distinct lines , m, and n concurrent at point C, choose
points L on and M on m distinct from C such that 0◦ < m∠LCM < 180◦ ;
let Θ ∈ m∠LCM. By Theorem 82,
ρC,2Θ = σ m ◦ σ .
(2.8)
−−→
Choose a point N on line n distinct from C and consider ray CN . There exist
−−→
−−→
−−→ −−→
unique rays CP and CQ such that the directed angle measure from CP to CN
−−→
−−→
←→
←→
and from CN to CQ equals Θ. So let p = CP and q = CQ (see Figure 2.11).
2.3. ROTATIONS AS PRODUCTS OF REFLECTIONS
m
p
l
Θ
n
q
49
Θ
C
Θ
Figure 2.11.
Then by Theorem 82 we have
ρC,2Θ = σ n ◦ σ p and ρC,2Θ = σ q ◦ σ n .
(2.9)
The result now follows from the equations in (2.8) and (2.9).
A translation τ is a product of two reflections in distinct parallels lines and
m, i.e., τ = σ m ◦ σ . Similarly, a rotation ρC,2Θ is a product of two reflections
in distinct lines and m intersecting at C, i.e., ρC,2Θ = σm ◦ σ . Furthermore, if
n, and m are concurrent at C, Theorem 84 tells us that n uniquely determines
lines p and q also concurrent with n at C such that σ m ◦ σ = σ n ◦ σ p = σ q ◦ σ n ,
where p is the unique line through C such that the measure of the directed angle
from p to n is Θ; q is the unique line through C such that the measure of the
directed angle from n to q is Θ. Thus:
Corollary 85 Let 0◦ < Θ◦ < 180◦ and let n be an arbitrarily chosen line
passing through point C. If Θ ∈ Θ◦ , there exist unique lines p and q passing
through C such that
ρC,2Θ = σ n ◦ σ p = σ q ◦ σ n .
Finally, if we multiply both sides of the equation σ m ◦ σ = σ n ◦ σ p on the
left by σ n we obtain σp = σn ◦ σm ◦ σ . Hence:
Corollary 86 Let , m, and n be distinct lines concurrent at point C. There
exists a unique line p passing through C such that
σp = σn ◦ σm ◦ σ ,
50
CHAPTER 2. CONGRUENCE
i.e., the product of three reflections in concurrent lines a reflection in some
unique line concurrent with them.
Corollary 87 A halfturn ϕP is the product (in either order) of two reflections
in perpendicular lines intersecting at P.
Note that the identity ι = τ PP = ρC,0 = σ ◦σ for all P, C, and . Therefore
Theorem 88 A product of two reflections is either a translation or a rotation;
only the identity is both a translation and a rotation.
Exercises
1. Lines
and m have respective equations X = 3 and Y = X.
a. Find the equations of the rotation ρC,Θ = σ m ◦ σ .
b. Find the center and angle of rotation C and Θ◦ .
2. Lines
and m have respective equations Y = X and Y = −X + 4.
a. Find the equations of the rotation ρC,Θ = σ m ◦ σ .
b. Find the center and angle of rotation C and Θ◦ .
3. Find the equations of lines
4. Let C =
and m such that ρO,90 = σ m ◦ σ .
£3¤
4 . Find equations of lines
and m such that ρC,60 = σ m ◦ σ .
5. Lines , m and n have respective equations X = 0, Y = 2X and Y = 0.
a. Find the equation of line p such that σ m ◦ σ = σ p ◦ σ n .
b. Find the equation of line q such that σ m ◦ σ = σ n ◦ σ q .
6. Let C be a point on line . Prove that σ ◦ ρC,Θ ◦ σ = ρC,−Θ .
7. If σp = σ n ◦ σ m ◦ σ , prove that lines , m and n are either concurrent or
mutually parallel.
8. If lines , m and n are either concurrent or mutually parallel, prove that
σn ◦ σm ◦ σ = σ ◦ σm ◦ σn.
2.4. THE FUNDAMENTAL THEOREM
51
9. Construct the following in the figure below:
b
a l
n
m
c
a. Line s such that σs = σ n ◦ σ m ◦ σ .
b. Line t such that σ t = σ c ◦ σ b ◦ σ a .
c. The fixed point of σ t ◦ σ s .
10. Given distinct points P and Q, construct the point R such that τ PQ ◦
ρP,45 = ρR,45 .
11. Let ¤ABCD ∼
= ¤EF GH be a pair of congruent rectangles. Describe how
to find a rotation ρP,Θ such that ρP,Θ (¤ABCD) = ¤EF GH.
12. Given distinct points P, Q and R, construct the point S such that τ QR ◦
ρP,120 = ρS,120 .
13. If , m and n are the respective perpendicular bisectors of sides AB, BC
and CA of 4ABC, find the line p such that σ p = σ n ◦ σ m ◦ σ .
2.4
The Fundamental Theorem
In this section we observe that every isometry is a product of three or fewer
reflections. This important result will be obtained by carefully analyzing the
set of points fixed by an isometry.
Theorem 89 An isometry that fixes two distinct points is either a reflection or
the identity.
←→
Proof. Let P and Q be distinct points and let m = P Q. We know that the
identity ι and the reflection σ m are isometries that fix both P and Q. Are there
any other isometries with this property? Let α be a non-identity isometry that
fixes both P and Q, and let R be any point not fixed by α. By Theorem 73,
52
CHAPTER 2. CONGRUENCE
R is off line m and P, Q, and R are non-collinear. Let R0 = α(R); since α is
an isometry, P R = P R0 and QR = QR0 . Consider the two circles PR and QR ,
which intersect at points R and R0 (see Figure 2.12).
R '=α(R)
P
m
Q
PR
R
QR
Figure 2.12.
Since P and Q are equidistant from R and R0 , line m is the perpendicular
bisector of chord RR0 . Hence
α(R) = R0 = σm (R),
α(P ) = P = σ m (P ),
and α(Q) = Q = σ m (Q)
and by Theorem 75,
α = σm .
Theorem 90 An isometry that fixes exactly one point is a non-identity rotation.
Proof. Let α be an isometry with exactly one fixed point C, let P be a
point distinct from C, and let P 0 = α(P ). Let m be the perpendicular bisector
of P P 0 (see Figure 2.13). Since α is an isometry, CP = CP 0 . Hence C is on m
and σm (C) = C. By the definition of a reflection, σ m (P 0 ) = P. Therefore
(σ m ◦ α)(C) = σ m (α(C)) = σ m (C) = C
and
(σ m ◦ α)(P ) = σ m (α(P )) = σ m (P 0 ) = P
so that σ m ◦α is an isometry that fixes two distinct points C and P . By Theorem
←→
89, either σ m ◦ α = ι or σm ◦ α = σ where = CP .
2.4. THE FUNDAMENTAL THEOREM
53
P
m
C
P'
l
Figure 2.13.
Lines
and m are distinct since P 6= P 0 . However, if σ m ◦ α = ι, then σ m = α,
which is impossible because α fixes exactly one point while σm fixes infinitely
many points. Therefore we are left with
σm ◦ α = σ
and hence
α = σm ◦ σ .
The fact that α is a non-identity rotation follows from Theorem 83.
In summary we have:
Theorem 91 Every isometry with a fixed point is either the identity, a reflection or a rotation. An isometry with exactly one fixed point is a non-identity
rotation.
Proof. By Theorem 89, an isometry with two (or more) distinct fixed points
is the identity or a reflection. By Theorem 90 and Theorem 83, an isometry
with exactly one fixed point is a rotation.
We are ready for our first of three important results in this course:
Theorem 92 (The Fundamental Theorem of Transformational Plane
Geometry): A transformation α : R2 → R2 is an isometry if and only if α
factors as a product of three or fewer reflections.
Proof. By Exercise 1.1.3, the composition of isometries is an isometry. Since
reflections are isometries, every product of reflections is an isometry. Conversely,
let α be an isometry. If α = ι, choose any line and write ι = σ ◦ σ , in which
case the identity factors as a product of two reflections. So assume that α 6= ι
and choose a point P such that P 0 = α(P ) 6= P. Let m be the perpendicular
bisector of P P 0 and observe that
(σ m ◦ α)(P ) = σ m (α(P )) = σ m (P 0 ) = P,
54
CHAPTER 2. CONGRUENCE
i.e., β = σ m ◦α fixes the point P. By Theorems 89 and 90, β factors as a product
of two or fewer reflections, which means that
α = σm ◦ β
factors as a product of three or fewer reflections.
Theorem 92 tells us that a product of eight reflections, say, can be simplified
to a product of three or fewer reflections. But how would one actually perform
such a simplification process? One successful strategy is to consider three noncollinear points and their images and follow the procedure that appears in the
proof of the next important theorem:
Theorem 93 4P QR ∼
= 4ABC if and only if there is a unique isometry α
such that α(P ) = A, α(Q) = B and α(R) = C.
∼ 4ABC, Theorem 75 tells us that if an isometry
Proof. Given 4P QR =
α with the required properties exists, it is unique. Our task, therefore, is to
show that such an isometry α does indeed exist; we’ll do this by constructing
α explicitly as a product of three isometries α3 ◦ α2 ◦ α1 each of which is either
the identity or a reflection.
Q
R
P
B
A
C
Figure 2.14.
Begin by noting that
AB = P Q,
AC = P R,
and BC = QR,
(2.10)
by CP CT C (see Figure 2.14).
The isometry α1 : If P = A, let α1 = ι. Otherwise, let α1 = σ where
is the
2.4. THE FUNDAMENTAL THEOREM
55
perpendicular bisector of P A. In either case,
α1 (P ) = A.
Let
Q1 = α1 (Q) and R1 = α1 (R)
and note that
P Q = AQ1 ,
P R = AR1 ,
and QR = Q1 R1
(2.11)
(see Figure 2.15).
Q
R
P
R1
l
B
α1(P)=A
C
Q1
Figure 2.15.
The isometry α2 : If Q1 = B, let α2 = ι. Otherwise, let α2 = σ m where m is the
perpendicular bisector of Q1 B. By (5.2) and (5.3) we have
AB = P Q = AQ1
so the point A is equidistant from points B and Q1 . Therefore A lies on m and
in either case we have
α2 (A) = A and α2 (Q1 ) = B.
Let
R2 = α2 (R1 )
and note that
AR1 = AR2
(see Figure 2.16).
and Q1 R1 = BR2
(2.12)
56
CHAPTER 2. CONGRUENCE
Q
R
P
R1
l
α2(Q1)=B
m
α2(A)=A
C
Q1
Figure 2.16.
The isometry α3 : If R2 = C, let α3 = ι. Otherwise, let α3 = σ n where n is the
perpendicular bisector of R2 C. By (5.2), (5.3), and (2.12) we have
AC = P R = AR1 = AR2
so the point A is equidistant from points C and R2 and lies on n. On the other
hand, (5.2), (5.3), and (2.12) also give
BC = QR = Q1 R1 = BR2
so the point B is equidistant from points C and R2 and also lies on n. In either
case we have
α3 (A) = A,
α3 (B) = B,
and α3 (R2 ) = C.
Let α = α3 ◦ α2 ◦ α1 and observe that
α(P ) = α3 (α2 (α1 (P ))) = α3 (α2 (A)) = α3 (A) = A
α(Q) = α3 (α2 (α1 (Q))) = α3 (α2 (Q1 )) = α3 (B) = B
α(R) = α3 (α2 (α1 (R)) = α3 (α2 (R1 )) = α3 (R2 ) = C.
Therefore α is indeed a product of three or fewer reflections. The converse
follows from the fact that isometries preserve lengths and angles.
The following remarkable characterization of congruent triangles is an immediate consequence of Theorem 93:
Corollary 94 4P QR ∼
= 4ABC if and only if 4ABC is the image of 4P QR
under three or fewer reflections.
2.4. THE FUNDAMENTAL THEOREM
57
Corollary 95 Two segments or two angles are congruent if and only if there
exists an isometry mapping one onto the other.
Proof. Two congruent segments or angles are contained in a pair of congruent triangles so such an isometry exists by Theorem 93. Since isometries
preserve length and angle the converse also follows.
Now we can define a general notion of congruence for arbitrary plane figures.
Definition 96 Two plane figures s1 and s2 are congruent if and only if there
is an isometry α such that s2 = α (s1 ).
Exercises
£¤
£¤
£0¤
£¤
£1¤
£ 12 ¤
1. Let A = 00 ; B = 50 ; C = 10
; D = 42 ; E = −2
; F = −4
. Given that
4ABC ∼
= 4DEF, find three or fewer lines such that the image of 4ABC
under reflections in these lines is 4DEF.
£¤
£3¤
£8¤
2. Let A = 67 ; B = 14
; C = 15
. In each of the following, 4DEF ∼
=
4ABC. Find three or fewer lines such that the image of 4ABC under
reflections in these lines is 4DEF.
a.
b.
c.
d.
e.
D
E
F
£12¤
£9¤
£14¤
1
£−1¤
10
£
¤
4
−9
£ −4 ¤
−15
£−5¤
−4
8
9
£−8¤
7
£
¤
7
−16
£−11¤
£−9¤
12
£
2
−17
¤
−12
£−12¤
−1
−6
£−12¤
−17
£−13¤
58
CHAPTER 2. CONGRUENCE