Math 347
#14. Prove the intersection of 2 convex sets is also convex.
Proof (Stephanie Paden): Let D and E be convex sets. Let A and B be elements of D E , so
that A and B are each in D and in E. Let C be a point on the segment AB with c  E D. This
would imply that c  E or c  D , which contradicts the fact that set E (set D, respectively) was
assumed convex. Therefore, the intersection of the sets is convex.
22. State and prove the angle-side-angle theorem.
Angle-Side-Angle Theorem: If two angles and the INCLUDED side of one triangle are
congruent to two angles and the INCLUDED side of another triangle, then the triangles are
congruent.
Proof (Martha Rojas):
Assume Postulate 15 (SAS Postulate.) Let A  EDF , B  E , and AB  DE . If BC  EF ,
then by SAS ABC  DEF and we are done. Assume BC  EF (note:   means is not
congruent to), then there exists a point G on the ray EF by F such that EG  BC . Hence, by
SAS, ABC  DEG . Since we assumed A  EDF and since ABC  DEG ,
A  EDG . Therefore, EDF  EDG . Thus, by problem # 7 DF and DG are the same
line segment. Therefore, since the segment can only intersect EG in one point, F and G are the
same point. Hence, EF  EG  BC . But this contradicts with our assumption that BC  EF .
Therefore, BC  EF , and so by SAS ABC  DEF .
23) Prove that if two angles of a triangle are congruent, the sides opposite these angles
are congruent.
Proof (Sharmaine Alexander):
Show that triangle BAC is congruent to triangle BCA. We are given that angle A is congruent to
angle C.
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We show that triangle BAC is congruent to BCA: Triangle BAC is congruent to itself in another
order. Angle A is congruent to angle C, angle C is congruent to angle A, and line AC is
congruent to line CA by the reflexive property. Since angle A is congruent to angle C and angle
C is congruent to angle A and line segment AC is congruent to line segment CA , by ASA
theorem, triangle BAC is congruent to triangle BCA.
Therefore the corresponding side BC is congruent to BA by definition of congruence.
QED.
24). Prove an equiangular triangle is equilateral.
<A  B  C
Fig 1.
Fig 2.
Proof (by Maria Rivas):
Since <A  <C, then ∆ABC is isosceles with base AC and equal sides AB  BC .
(by problem 23)
Similarly, since <A  <B then ∆ABC is isosceles with base AB and equal sides AC  CB .(by
problem 23)
Since AB  BC and AC  CB , then by the transitive property AB  BC  AC .
Therefore all sides are congruent. Hence an equiangular triangle is equilateral.//
25. State and Prove the theorem of side-side-side triangle congruence.
Side-side-side (SSS)
If three sides of one triangle are congruent to three sides of another triangle, then the triangles are
congruent.
Show that If AB  DE , AC  DF and BC  EF , then the two triangles are congruent.
Proof (Erika Rey):
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Suppose that A  EDF , then the two triangles are congruent by SAS, and we are
done.
Now we will prove by contradiction that A  EDF .
Assume A  EDF
F
J
H
D
F
J
H
E
If we draw a line DG ,so that BAC  EDG .There exist a point H on the same side of
the ray EF , such that DH  AC. If F and H are not the same point then DF and DG can
only intersect at D. If we draw EH and FH then the ABC  DEH by SAS. This
implies that EH  BC by the definition of congruent triangles. Also EH  BC  EF
and DH  AC  DF . Both DFH and EFH are isosceles triangles by the definition of
isosceles triangles. If we take the point J to be the midpoint of FH , then the angle
bisector of FDH contains JD and the angle bisector of FEH also contains JE by
problem 28. This implies that D must be equal to E. This contradicts our assumption
that A  EDF because D can not be equal to E. Therefore A  EDF , and the
triangles are congruent by SAS.
Problem #26:
State and prove the theorem of Side-Angle-Angle triangle congruence.
SAA: If two angles and a side of the first triangle are congruent to the corresponding parts of
the second, then the triangles are congruent.
B
A
E
C
D
G' F
G"
Let the given SAA correspondence be: <A is congruent to <D, <C is congruent to <F, and
segment AB is congruent to segment DE.
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Proof (Bolivia Brandt): If segment AC is congruent to DF, then ∆ ABC is congruent to ∆ DEF
by ASA and we are done.
We need to prove that segment AC is congruent to segment DF.
Let point G be a point on ray DF such that DG = AC.
Then we have that ∆ ABC is congruent to ∆ DEG by SAS. By corresponding parts, we have
<EGD is congruent to <C congruent to <EFD.
There are two cases: either (1) DF>DG, shown as G=G’, or (2) DF<DG, shown in the diagram
as G=G’’. In case (1), <EG’D is an exterior angle of ∆ EFG’ and <EG’D is congruent to <EFD,
a remote interior angle of the triangle; in case (2), <EFD is an exterior angle of ∆ EFG” and
<EFD is congruent to <EG’’D, a remote interior angle of the triangle. In either case, by the
exterior angle theorem (38), the exterior angle must be strictly larger than either remote interior
angle, and so we have a contradiction.
Therefore, point F = G. Hence ∆ ABC is congruent to ∆ DEF by ASA.
27A. Is it possible for two lines to intersect in
such a way that three angles formed measure 30, 80,
and 30?
No:
B
|
|
A-------X-------C
|
|
D
Given: line AB and line CD intersect at point X.
Proof (Joy):
If <AXB and <CXB are a linear pair, then they are supplementary. (postulate 14). By definition,
<AXB and <CXB share a common ray XB, and A, X and C are collinear, then <AXB and <CXB
are supplementary.
Since <AXB and <CXD are vertical angles, based on Problem
16, <AXB is congruent to <CXD.
If <AXB = 30 then <CXB must equal 150.
If <AXB = 30 then <CXD must equal 30.
Then <BXC and <AXD must both = 150 not 80.
27B. Carefully state the conditions under which it is impossible for two lines to intersect in four
angles of measures a, b, c and d.
Solution (Joy): Conditions under which the angles are possible:
a=c, d=c, a+b=180, d+c=180.
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Conditions which are impossible:
If any of the following inequalities holds, then it is impossible to construct two lines with the
given angle measures:
a+b does not equal 180
d+c does not equal 180
a does not equal c
d does not equal c
C
E
A
D
B
28. Proof (Ryan Flores): Let E be any point on CD , the perpendicular bisector of
given segment AB . Let A and B be the given endpoints of AB .
By the definition of perpendicular bisector, AD  DB and EDB  EDA . Also,
ED  ED by the Reflexive Property.
 By SAS (Postulate 15) EDB  EDA .
 AE = BE by definition of congruent triangles (corresponding parts), and AE=BE
(definition of congruent segments).
 Any point E on the perp. bisector CD is equidistant to the endpoints of AB .
29.
Prove that given a point on a line in a plane, there is exactly one line perpendicular to the
given line through that point.
Given: A point on a line in a plane,
Prove (Keith Whitson): There is exactly one line perpendicular to the given point through a line
in the plane.
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W
R
J
T
L
Q
Existence: Let R be a plane and let T, Q be points that determine a line in that plane. Let L be
the given point on line TQ. Let W be a point not on line TQ to create a ray LW such that angle
QLW is a right angle by the angle construction postulate. Therefore, ray LW is perpendicular to
line TQ, and so is line LW.
Uniqueness: Pick another point J not on line TQ such that angle QLJ is a right angle by the
angle construction postulate and so line LJ is perpendicular to line TQ. This is a contradiction to
the angle construction postulate because the two angles much have the same endpoint because
they have because they have the same angle measure so they must be the same angle and so the
line is unique.
Therefore, given a point on a line in a plane, there is exactly one line perpendicular to the given
point through a line in the plane.
32.) Ryan Flores
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B
A
D
C
Let ABC be a  having side AC  AB.
Prove  ABC   ACB.
Since AC  AB, make AD  AB and join BD.
Since ADB is an exterior  of the BCD, it is greater than the interior and opposite DCB.
But ADB  ABD, since the side AB  AD.
 ABD  DCB=ACB.
 ABC  ACB.
 In any , the  opposite the greater side is greater.
Problem 33 can be obtained by emailing Prof. Jones at mjones@csudh.edu. It is a large JPEG
file.
34. Prove the triangle inequality: in any triangle ABC, AB + BC is greater than or equal
to AC.
Proof (by Mitsuyo Clark)
Let ABC be any triangle. Construct a point D on the line AB such that A-B-D and
BC = BD.
Then AD = AB + BD by definition of three points. ---(1).
 ACD =  ACB +  BCD by Angle Addition Postulate. Hence  ACD >  ACB and
 ACD >  BCD. ---(2)
Since the triangle BCD is isosceles with BD = BC, then  BDC is congruent to  BCD. --- (3)
From (2) and (3),  ACD >  BDC by transitive property. Then by problem 33, the side AD is
greater than the side AC. Hence AD > AC in terms of distance. ---(4)
From (1) and (4), AB + BD = AD > AC, and BD = BC, by transitive property, we have AB + BC
> AC.
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C
D
A
B
35. Prove that if two lines in a plane are both perpendicular to the same line, then they are
parallel.
A
l
B
m
n
Given: Line l is perpendicular to line n at point A.
Line l is perpendicular to line n at point B.
Prove: Line l is parallel to line m.
Proof (Joel Negrete): We will assume that the two lines (l and m) are not parallel
and will intersect at a point P.
l
A
B
P
m
n
If the two lines intersect then we have two perpendicular lines to line n through
point P which is a contradiction (proved in problem 29).
Therefore lines l and m are parallel.
36.
Prove that there is at least one line through a given point not on a given line that is
parallel to the given line.
P
n
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l
m
Given: A line l and a point P not contained in line l
Prove: we can construct a parallel line to line l through point P.
Proof (Alejandro): Given a line and a point not on the given line we can find a plane that
contains both the line and the point.
By Postulate 7: There is at least one plane containing any three given points. If the points
are non-collinear, then there is only one such plane.
Since both the line l and the point P are on the plane S, we can construct a perpendicular line m
through point P to line l.
By problem # 31 - given a point not on a given line in a plane, there is exactly one line
perpendicular to the given line.
Since point P is contained in line m and line m is contained in the same plane as point P and line
l, we can construct a line n perpendicular to line m through point P which is contained in line m
By problem # 29 - given a point on a line in a plane, there is exactly one line perpendicular
to the given line through that point.
Since both line l and line n are perpendicular to the same line m, then line l is parallel to line n.
By problem 35, if two lines in a plane are both perpendicular to the same line, then they are
parallel.
Homework Problem # 37
a) Define alternate interior angles.
When two lines are crossed by a transversal, two angles located inside two
lines and on opposite sides of the transversal are called alternate interior
angles.
b) Prove that if two lines are cut by a transversal, and if one pair of alternate
interior angles is congruent, then the other pair of alternate interior angles
is also congruent.
Given: m1  m3
Prove: m2  m4 (Susana Granados)
Statements/Conclusion
1) m1  m3
2) m1 + m2 = 180
Justification
1) Given
2) Supplementary angles.
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m3 + m4 = 180
3) m1 + m2 = m3 + m4
4) m1 + m2 = m1 + m4
5) m1 + m2 = m1 + m4
- m1
- m1
6) m2 = m4
Supplementary angles.
3) Set equal to each other.
4) Given, Substitute m1 with m3
5) Subtract m1 from both sides
1 2
4 3
Problem # 40 presented by Juan Castillo
Two nonadjacent angles made by the crossing of 2 lines by a third line, one angle being interior,
the other exterior, and both being on the same side of the line.
1
3
5
2
l
4
6
m
7 8
t
It is given that angle 2 is congruent to angle 6. We want to prove that angle 1 is congruent to
angle 5, angle 7 is congruent to angle 3, and angle 8 is congruent to angle 4.
Proof:
By the problem on vertical angles (16(b)), angle 2 is congruent to angle 3 and angle 6 is
congruent to angle 7. Therefore, angle 3 is congruent to angle 7 by substitution or transitive
property. By definition of supplementary angles: the sum of the measurements of angle 2 and
angle 1 is 180 degrees. As a result, the measurement of angle 2 is equal to 180 degrees minus
the measurement of angle 1. Similarly, the sum of the measurements of angle 6 and angle 5 is
equal to 180 degrees. As a result, the measurement of angle 6 is equal to 180 degrees minus the
measurement of angle 5. Since angle 2 is congruent to angle 6, then 180º - m<1 = 180º - m<5.
Therefore, the measurement of angle 1 is equal to the measurement of angle 5, which implies
that angle 1 is congruent to angle 5. By definition of vertical angles angle 1 is congruent to
angle 4 and angle 5 is congruent to angle 8. Since angle 1 is congruent to angle 5, then angle 8
is congruent to angle 4 by the transitive property.
41. Prove that if two parallel lines are cut by a transversal, then alternate interior angles are
congruent.
Math 347
Proof (Stephanie Paden): Below is my proof for #41. It was determined by the class and Dr.
Jones that my reason for statement #2 could not be ‘by problem #40’, since we are not given a
set of corresponding angles.
t
2 1
4 3
a
5 6
8 7
b
Statements
Reasons
1. a || b; t is transversal
1. Given
2. 2  6 and 1  5
2. By Problem #46 & #48. (corr
3. 4  2 and 3  1
3. Vertical s
4. 4  6 and 3  5
4. Transitive Property
s)
For reason #2, problem #48 asks for angles created when 2 parallel lines crossed by a third line.
Prof. JONES’ note: #46 is not a valid reason for line 2. #48 is a valid reason, but then you need
to prove some other set of angles, such as corresponding angles, is congruent. This is a long way
to get to #41. To prove #41 without using #48, use proof by contradiction: Let the intersection
of the transversal with line b be point B. If angles 2 and 6 are not congruent, then there is a line
b’ with angle 6’ at B congruent to angle 2. Then by #39, b’ is also parallel to a. But by postulate
16, there can only be one line through B parallel to a. This is a contradiction. Thus the alternate
interior angles 2 and 6 are congruent, and by 37b, angles 1 and 5 are also congruent.//
Problem 42
Let the line M be the transversal to the lines L and N
Also L // N
M
N
L
M
P
1
3
Proof (Brahim Karam):
(1 ) Then by theorem 41, we have m< 1 = m<3
K
L
P
2
3
(alternate interior angles)
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Now, let’s assume that there exist another line K through P such that:
K // L and m< 2 ≠ m< 1
(2 ) Then, by theorem 41, we have m< 2 = m< 3 (alternate interior angles)
(3) therefore, by (1) and (2) we have m< 2 = m< 1 (transitivity)
By statement (3) is a contradiction to the statement m< 2 ≠ m< 1
Therefore, there is only one line that is parallel to L through the point P
43. A close translation of the Parallel Postulate as Euclid gave it is: If a third line crosses two other lines, and
the interior angles on the same side of the third line have an angle sum less than 180 degrees, then the lines,
if extended sufficiently, meet on the side in which the angle sum is less than 180 degrees.
a. Restate the postulate in your own words.
b. Prove Euclid’s version as a consequence of Postulate 16.
Solution (Ryan Flores, with substantial revision by Prof. Jones):
a)
Postulate 16 (Parallel Postulate) – If given 3 non-collinear points ABC, and if a line passes through points
Aand B, then there exists a unique line through point C parallel to the given line through A and B.
b)
Let m be a line and let o be another line with transversal line p, and with a same-side interior angle sum of
less than 180. Let o intersect p at point A. We need to show that o intersects m on the side on which the angle sum
is less than 180. Construct a transversal line n perpendicular to line m and passing through A (possible by 31), and
construct a line l perpendicular to n and passing through A (possible by 29). By problem 36, line l is parallel to line
m. Because we know that l || m, we can say that m1 = 90, m2 = 90, m3 = 90, and m4 = 90. We know
this by the results of problems 41 and 16. Therefore, m1 + m3 = 180 and m2 + m4 = 180.
n
A
1
2
3
4
l
m
B
Now, line o also passes through point A. By Post. 16, we know there cannot be more than one line || to m through
point A. Knowing this, we can say that if o WERE || to m through A then o and l must be the same line. Thus o is
not parallel to m and the lines meet.
B
p
m
A
o
l
It remains to show that o and m meet on the side in which the same-side interior angle sum is less than 180.
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Since l and m are parallel, and p is a transversal, we know that the same-side interior angle sum is equal to 180 on
both sides (by 41 and linear pairs). Let B be a point on line o on the side of p in which the sum is greater than 180.
By plane separation, and angle addition, ray AB is on the opposite half-plane from line m, and hence does not
intersect line m. Thus the lines must meet on the side in which the sum is less than 180.
Problem 44
A-Let the line M be the transversal to the lines L and N. Also L // N
M
Q
N
L
3
P
4
11’
2
Proof (Brahim Karam):
It follows by Theorem 43 that <1 and <2 are supplementary, as are <3 and <4. Let Q be another
line // to L through the point P, such that
First case:
m< 1’ is less that m<1.
Then, the sum of m< 1’ and m< 2 is less than 180 degrees.
Also, <1’ and <2 are on the same side of the transversal M
Therefore, by theorem 43 the line Q and the line L are not parallel, which contradicts the
given that Q //L
Second case:
m<1’ is bigger than m<1
Then the sum of m<3 and m<4 is less than 180 degrees
Hence, again by theorem 43 Q is not parallel to L
In both cases, we have if a line through the point P and parallel to the line L then it is unique.
B- (Brahim)
Yes, theorem 43, stated as, “If a third line crosses two other lines, and the interior angles on the
same side of the third line have an angle sum less than 180 degrees, then the lines, if extended
sufficiently, meet on the side in which the angle sum is less than 180 degrees,” and which is in
the form ┐Q┐P, is equivalent to the statement, “If two parallel lines are cut by a transversal,
interior angles on the same side of the transversal are supplementary,” which is in the form of
P Q. These two forms are logically equivalent and one is the contrapositive of the other.
45.
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D
B
E
A
C
Prove: m<A +m<ABC+m<C = 180 degrees (Edgar E. Perez)
1.)
2.)
3.)
4.)
Line DE ll line AC by construction
m<DBE = 180 degrees by linear pair supplementary definition
m<DBA + m<ABC+m<CBE = 180 degrees by linear pair/supplementary
m<A = m<DBA
By #47 alternate interior angles of parallel lines are congruent
m<C = m<CBE
Hence, m<A+m<ABC+m<C = 180 degrees by transitivity
49. Show that the image of a segment under an isometry is also a segment.
A
C
A'
C'
B
B'
Proof (Martha Rojas):
We know that a transformation T is an isometry if it preserves distance.
Let T be any isometry and AB be any segment with T(A) = A’ and T(B) = B’. Because T is an
isometry, AB = A’B’. Therefore, we only need to show that T ( AB)  A' B' , that is, show that a
point C is on AB if and only its image T(C) = C’ is a point on A' B' . Since we know T(A) = A’
and T(B) = B’, we only need to consider points C and C’ between those points. To show
T ( AB)  A' B' , we need to show two things: 1) T ( AB)  A' B' and 2) A' B'  T ( AB) .
To show that T ( AB)  A' B' , suppose C is on AB and between A and B. Then, AC + CB = AB
by definition of between. Since T preserves distance, A’C’ = AC and C’B’ = CB. Therefore,
A’C’ + C’B’ = A’B’. This implies that C’ is between A’ and B’ and thus C’ is on A' B' .
To show that A' B'  T ( AB) , we do the opposite. Suppose C’ is on A' B' and between A’ and
B’. Then, A’C’ + C’B’ = A’B’. Again, since T preserves distance, it must be that its preimage C
satisfies AC + CB = AB. Therefore, C is between A and B and on AB .
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Therefore, T ( AB)  A' B' or the image of a segment under an isometry is also a segment. 
PROBLEM #50
Show that the image of a ray under an isometry is also a ray.
Proof (Bolivia Brandt): From the definition of isometry and from 49, we know: Every isometry
preserves angle measure, betweennes, collinearity, and distance (length of segments).
Suppose we are given ray XY, and that the images of points X and Y are X’ and Y’.
Y
Y'
B
B'
A
A'
X
X'
By problem #49: A segment under an isometry is also a segment. A ray contains an infinite
amount of segments and each segment is preserved under the isometry; in other words, for each
segment XZ of ray XY, segment X’Z’ is a corresponding and congruent segment of ray X’Y’.
Therefore, we can say that any other point in the ray XY will be on the ray X’Y’ according to the
above definition. Hence, the image of a ray under an isometry is also a ray.
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51. Show that the image of an angle under an isometry is also an angle.
Given:  PQR
Prove:  P’Q’R’
Proof: (Erika Rey)
P
Q'
Q
P'
R
R'
QP is isometric/congruent to Q' P' and QR is isometric/congruent to Q' R' by problem 50. Since
both Q' P' and Q' R' share Q’, by angle construction two nonlinear rays with a common vertex
form an angle. Therefore the image of an angle under an isometry is also an angle. QED
52. Show that segments congruent under the new definition are congruent under the old
definition.
Solution (Keith Whitson):
Old Definition: Two segments are congruent if the have the same length.
New Definition: Two objects in a plane are said to be congruent if there is an isometry with
T(A)=B.
Isometry: A transformation is an isometry if it preserves distance.
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By #49, the image of a segment is also a segment. Let T be an isometry and segment AB be any
segment with T(A)=A’ and T(B)=B’. Because AB=A’B’ under the new definition, which
preserves distance, segment AB is congruent to segment A’B’ under the old definition because
the two line segments have the same length.
53) Show that congruent angles under the new definition are congruent under the old definition.
*OLD DEFENITION - Two angles are congruent if their measures are equal.
*NEW DEFENITION – Two objects A and B in a plane are said to be an isometry with T(A)
= B.
***Show that T(< PQR) is congruent to (< P’Q’R’)****
Proof (Belinda Vargas):
Assume P’, Q’, and R’ are the images of P, Q, and R, respectfully, under the isometry. Since
isometries preserve distance, we know PQ=P’Q’, PR=P’R’, and QR=Q’R’.
Therefore, triangle PQR is congruent to P’Q’R’ (by SSS). So, by corresponding parts < PQR
is congruent to < P’Q’R’ (by old definition)
P
P'
Q
Q'
R
R'
54. Assume that two triangles are congruent under the new definition. Then
there is an isometry which maps one triangle onto the other. Show that
these triangles are congruent under the old definition.
Proof (by Mitsuyo Clark)
Given an isometry T with T(ΔABC) = ΔA'B'C' and ΔABC  ΔA'B'C' under the new
definition. Since distance is preserved under an isometry T, AB = A'B', BC
= B'C', and AC = A'C'. By problem 52 (segments congruent under the new
definition are congruent under the old definition), AB  A' B ' , AC  A'C ' , and BC  B'C ' . By
SSS theorem, ΔABC  ΔA'B'C' under the old definition.
Problem 55 can be obtained by emailing Prof. Jones at mjones@csudh.edu. It contains scanned
images in a word document.
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56). Prove that rotations are isometries.
Given: A’ and B’ are the rotation images of A and B after a rotation about C.
Prove: AB  A' B '
Proof: (Maria Rivas): By cases.
Case 1. A=C
Since A=C, AB=CB. By definition of rotation CB=CB’. By transitivity, AB=A’B’.
(AB=CB=CB’=A’B’). Similarly if B=C.
Case 2. ABC are collinear but distinct.
By segment addition, AB= AC+CB and A’B’= A’C+CB’. By definition of rotation CB= CB’
and CA=CA’. By transitivity, AB=A’B’.
Case 3. A, B ≠ C and noncollinear
Math 347
By definition of rotation, CA=CA’, CB=CB’, and  ACA’  BCB' .
By definition of congruence, m ACA'  mBCB ' , hence m 1 + m 2 = m  ACA’ and m  2+
m  3= m  BCB’. By transitivity, m 1 + m 2 = m  2+ m  3, it follows that, m 1 = m  3.
By SAS theorem, ∆ABC  ∆A’CB’. Therefore, by corresponding parts of congruent triangles,
AB=A’B’.
Math 347
#57) Prove that translations are isometrics.
Proof (Sarah O’Crowley): We must show that the distance between two given points J
and K is preserved under translation.
Case 1 :
If one pair of opposite of a quadrilateral is parallel and congruent, then the
quadrilateral is a parallelogram. JJ’ and KK’ are parallel and congruent, so JJ’KK’ is a
parallelogram. Since the opposite sides of a parallelogram are congruent, JK is congruent
to J’K’ and JK = J’K’.
J’
J
K
K’
Case 2: When JK is moved along a vertical line
It follows from the definition of translation that JJ’=KK’.
1) K’K = K’J’ + J’K
2) JJ’ = J’K + KJ
By substitution and cancellation, K’J’ =KJ.
K’ *
*J’
*K
*J
60. Prove that the composition of two isometries is also an isometry.
Note: A transformation T is an isometry if it preserves distance.
Proof (Sharmaine Alexander):
Let F and G be two isometries. That means F and G both preserve distance.
For any two given points, p and q, we have the following:
G (p) = p’ and G (q) = q’. Since G is an isometry we can say that pq = p’q’.
Also F (p’) = p” and F (q’) = q”. Since F is an isomtery we can say that p’q’ = p”q”.
Therefore pq = p’q’ = p”q”. Thus F ○ G (p) = p” and F ○ G (q) = q”.
Therefore the composition of F and G is an isometry.
Math 347
61. Hint: Showing congruence under the new definition means that one must describe an
isometry, or a sequence of isometries (allowed because of 54), in which the image of the
first object is the other object.
a. Prove that two segments congruent under the old definition are congruent under
the new definition.
Given: Segment AB is congruent to segment CD (old definition).
Proof (Keith Whitson): Show that there is a distance-preserving transformation that maps
segment AB onto segment CD. Let m be the perpendicular bisector of segment AC.
Then the reflection rm maps segment AB onto a segment A’B’, and A’=C by construction.
Let n be the bisector of angle B’CD. Because CD=AB=A’B’=CB’ and n is the bisector,
rn (segment CB’) = segment CD. Thus, rn○rm (segment AB) = segment CD, and so
segment AB is congruent to segment CD. QED.
n
A
C = A'
B'
D
B
m
Math 347
b. Prove that two angles congruent under the old definition are congruent under the
new definition.
Given: <ABC is congruent to <DEF (old definition)
Prove: Show that there is a distance-preserving transformation that maps <ABC onto
<DEF.
Proof: Let t be the translation that connects vertex B to vertex E, then let r be the
rotation by the angle A’B’D that rotates rays A’B’ and B’C’ into the positions of DE
and EF respectively (it may be necessary to reflect across the bisector of <DEF to get
proper alignment of the points D, E, and F with A, B, and C). Thus, r composed with
t (and perhaps a reflection) is an isometry that preserves the angle measurement of the
angles. Therefore, two angles congruent under the old definition are congruent under
the new definition.
C''=F
B
E=B'
A
A''=D
C
A'
C'
62). Prove that two triangles congruent under the old definition are congruent under the new
definition.
Old definition: Two triangles are congruent if there is a correspondence between the angles and
the segments of each triangle such that the corresponding angles and segments are congruent.
New definition: Two objects in the plane are said to be congruent (A congruent to B) If there is
an isometry T (A) =B.
Prove that ∆ABC  ∆ A’B’C’ (which are congruent under the old definition) are also congruent
under the new definition. In other words, there is an isometry U such that
U (∆ABC) = (∆ A’B’C’).
Math 347
Proof (Maria Rivas): Given two congruent triangles, wlog we can assume ∆ABC  ∆ A’B’C’
by SSS theorem. (Old definition). Therefore, AB = A’B’, BC= B’C’ and AC= A’C’.
Since AB =A’B’, there exists an isometry T such that T (AB) = A’B’.
Case 1). If T(C) = C’, then set U=T and the proof is done because every isometry preserves
distance.
Case 2). Because line A’B’ divides the plane in half and because T preserves angles (by 61b),
there are only two possibilities for the isometry: either T(C) = C’ or rA'B'  T (C )  C’ (which
can also be expressed as T(C) = C”). Since T (C) = C’ was proven in case 1, wma that the
isometry U= rA'B'  T .
Therefore U (∆ABC) = (∆ A’B’C’) and so ∆ABC  ∆ A’B’C’ under new definition.
Problem 63 (Susana Granados): The Pythagorean Theorem states that if ABC is a
right triangle with legs of length a and b, and a hypotenuse of length c, then a2 + b2
= c2. Prove the Pythagorean Theorem.
We start with four copies of the same triangle. Three of these have been rotated
90o, 180o, and 270o, respectively.
Each has area ab/2. Let's put them together without additional rotations so that
they form a square with side c.
Math 347
The square has a square hole with the side (a-b). Summing up its area (a-b)2 and
2ab, the area of the four triangles (4·ab/2), we get
c2 = (a - b)2 + 4 * (ab/2)
c2 = (a - b)2 + 2ab
c2 = a2 – 2ab + b2 + 2ab
c2 = a2 + b2. QED
64. Show that with SSA, there can be two different, incongruent triangles which both
satisfy SSA.
Solution (Ventress Sulcer):
SSA - If two sides and a non - included angle of one triangle are congruent to the corresponding
parts of another triangle ,then the triangles are congruent.
Given two triangles ABC and XYZ
The line AB is congruent to line XY, line AC ls congruent to line XZ and.angle C is
congruent with angle Z .(SSA) Even though two sides and two angles are congruent
triangles ABC and XYZ are not congruent.
66. What happens to the plane (or objects in the plane, if you prefer) if you perform two
reflections? Put another way, what is the result of performing 2 reflections across lines l1 and l2
in the plane? There are three possible cases for the two lines of reflection: l1 and l2 are
identical, they are parallel to each other, or they intersect.
Prove that if the lines are parallel, the result is a translation.
Math 347
Proof (Alejandro Vidal): If line m and line n are parallel, then rn ο rm is the translation with
direction perpendicular to the lines m and n, and with magnitude twice the distance between m
and n, in the direction from m to n.
Case 1
Case 2
A
P
A’
x
x
x
x A
A’y Q
y
y
2x + y
A”
m
n
m
n
A”
Proof
Suppose m and n are parallel lines. Let A’ = rm (A) and let A” = rn (A’) = rn ο rm (A). Since
segment AA’ and segment A’A” are perpendicular to parallel lines, the three points A, A’, and
A” are collinear. So line AA’’ is perpendicular to line m.
To find the length of AA”, notice that lines m and n are perpendicular bisectors of segment AA’
and segment A’A”. Consequently, if the distance between lines m and n is x + y as show above,
then AA” = 2x + 2y.
rn ο rm = TAA”  AA” = 2(PQ)
Segment AA” is perpendicular to lines m and n. Thus the composition of the two reflections is
the translation TAA’’, where the translation is perpendicular to lines m and n in the direction from
m to n, and is by a distance twice that of m to n.
Note: rn ο rm ≠ rm ο rn .