Download MAT 211: Final Exam Review Student‐Written Questions Fall 2010

MAT
211:
Final
Exam
Review
Student‐Written
Questions
Fall
2010
Chapter
1
(Intro)
1. A
triangle
is
isosceles
if
and
only
if
the
base
angles
are
congruent.
(Katy
Spencer)
2. Prove
that
the
diagonals
of
a
parallelogram
bisect
one
another
(relies
on
Euclid’s
5th
Postulate)
(Maggie)
3. A
quadrilateral
is
said
to
be
cyclic
if
there
is
a
circle
passing
through
all
its
four
vertices.
Given
a
quadrilateral
is
cyclic,
prove
that
the
sum
of
the
measures
of
its
non‐adjacent
interior
angles
is
180
degrees.
(Greg)
Finite
geometry
(Fe‐Fo)
4. Given
the
following
set
of
axioms,
write
three
conjectures
and
prove
one
of
them.
(Ryan
D.)
Axiom
1:
There
exist
exactly
n
points
in
the
system.
Axiom
2:
Every
set
of
two
points
lies
on
exactly
one
line.
Axiom
3:
Each
line
contains
exactly
two
points.
5. Prove
the
conjecture
from
the
given
axioms.
(Stephanie
M.)
A1:
There
is
a
group
of
five
athletes.
A2:
For
every
three
athletes,
there
is
a
common
sport
they
have
shared.
A3:
Not
all
of
the
athletes
have
played
the
same
sport.
Conjecture:
There
are
10
possible
sports
that
have
been
shared.
6. There
are
many
different
theorems
that
can
be
proved
from
the
Axioms
below.
Prove
one
of
the
theorems
provided.
(Stephanie
S.)
Axiom
1.
There
exist
exactly
three
distinct
students,
and
three
distinct
colleges.
Axiom
2.
Two
distinct
students
belong
to
exactly
one
college.
Axiom
3.
Not
all
students
belong
to
the
same
college.
Axiom
4.
Any
two
distinct
colleges
contain
at
least
one
student
which
belongs
to
both.
Theorem
1.
Two
distinct
colleges
contain
exactly
one
student.
Theorem
2.
A
college
cannot
contain
three
distinct
students.
Theorem
3.
There
exists
a
set
of
two
colleges
that
contain
all
the
students.
Chapter
2
(Triangles)
7. Consider
the
triangle
ABC
with
incenter
I.
Prove
that
the
incenter
is
a
point
that
is
equidistant
from
each
side
of
the
triangle.
(Ellen)
8. Determine
whether
the
following
statements
are
true
or
false.
If
the
statement
is
an
implication
state
the
hypothesis
and
conclusion.
Then
state
the
converse
and
the
contrapositive.
(Toni)
1.
If
A
is
a
parallelogram
than
the
diagonals
bisect
each
other.
2.
If
A
is
a
right
triangle
than
it
is
always
an
isosceles
triangle
3.
The
center
of
a
circle
lies
on
the
perpendicular
bisector
of
two
points
4.
All
rhombi
are
squares
5.
Every
rectangle
has
three
sides,
and
all
right
triangles
are
equiangular.
9. Assuming
that
SAS
is
true,
prove
ASA.
(Anna)
10. Prove
that
any
X
on
the
perpendicular
bisector
of
segment
AB
is
the
center
of
a
circle
through
points
A
and
B.
(Ketty)
Chapter
3
(Circles)
11. Two
circles
are
said
to
be
orthogonal
if
their
tangents
are
perpendicular
at
their
point
of
intersection.
(Clarissa)
a.
Describe
how
to
construct
orthogonal
circles.
b.
Prove
that
there
are
perpendicular
tangents
at
the
other
point
of
intersection
of
the
circles.
12. Prove
or
Disprove
that
the
area
of
the
semi‐circles
on
the
legs
of
a
right
triangle
(AB
&
BC)
sum
to
the
area
of
the
semi‐circle
on
the
hypotenuse
(CA).
(Mike)
13. Given
triangle
XYZ,
extend
sides
XY
and
XZ
creating
exterior
angles
at
Y
and
Z.
Prove
that
the
bisectors
of
the
exterior
angles
at
Y
and
Z
are
concurrent
with
the
bisector
of
the
interior
angle
at
X.
Call
this
point
of
intersection
R.
(Caitlin)
14. Prove
that
the
opposite
angles
in
a
cyclic
quadrilateral
ABCD
are
supplementary.
(Joe
K.)
Chapter
4
(Analytic
geometry)
15. Prove the Pythagorean theorem using analytic geometry. Be sure to justify that the
points you
use
make
a
right
triangle.
(Adam)
16. Prove
the
midpoint
formula
(Kevin)
17. Prove
that
the
diagonals
of
a
rhombus
are
perpendicular
bisectors
of
each
other
(Katie
Skerik)
18. (Jamie)
a.)
Using
coordinates,
construct
a
right
triangle.
b.)
Find
the
midpoint
of
the
hypotenuse.
c.)
Prove
(using
the
coordinates)
that
the
midpoint
of
the
hypotenuse
is
equidistance
from
all
the
vertices
of
the
triangle.
Chapter
6
(Transformational
geometry)
19. List
the
orientation
and
fixed
points
if
any
for
the
following
composition
of
isometries.
(Justin)
1. Reflection
followed
by
a
translation
2. Reflection
followed
by
a
rotation
3. Rotation
followed
by
a
translation
20. When
composing
isometries
which
compositions
can
be
replaced
by
a
single
isometry?
(Joe
H.)
21. Sometimes
a
composition
of
two
reflections
has
fixed
points
and
sometimes
it
doesn’t.
Write
a
paragraph
in
which
you
identify
when
it
has
fixed
points
and
explain
why
this
happens.
(Ryan
F.)
Chapter
9
(Hyperbolic
and
Spherical
Geometry)
22. Answer
if
the
following
questions
are
true
always,
sometimes,
or
never
in
hyperbolic
geometry.
Justify.
(Kim)
1.
AB
is
a
superparallel
to
line
l.
AB
is
the
closest
possible
parallel
to
line
l.
2.
Line
j
is
perpendicular
to
line
k.
Line
k
is
perpendicular
to
line
l.
There
is
a
line
m
that
is
perpendicular
to
both
lines
j
and
l.
3.
Construct
arbitrary
triangle
ABC.
Point
D
is
located
on
line
AC
and
is
between
points
A
and
C.
The
defect
of
triangle
ABD
is
greater
than
the
defect
of
triangle
ABC.
23. Prove:
AAA
is
a
congruence
postulate
in
Hyperbolic
Geometry.
(Kaitlyn)
24. Construct
Saccheri
Quadrilateral
ABCD
and
the
triangle
associated
with
it
with
vertex
E
and
intersecting
the
base
at
points
F
and
G.
F
and
G
and
constructed
to
be
midpoints
of
AE
and
BE
respectively.
Prove
the
angle
sum
of
a
triangle
is
equal
to
the
sum
of
the
summit
angles
of
its
associated
Saccheri
Quadrilateral.
(Lina)
25. True
or
False
–
provide
justification
using
the
Upper
half
plane
model.
(Dan)
a. Given
h‐line
A
and
point
B,
not
on
A,
there
exists
one
and
only
one
h‐line
parallel
to
A.
b. If
C
and
D
are
two
parallel
h‐lines,
and
D
is
parallel
to
h‐line
E
as
well,
then
C
and
E
are
also
parallel.
c.
Given
an
h‐line
A
and
a
point
B,
not
on
A,
there
exists
one
unique
perpendicular
through
A
to
B.
26. State
whether
the
following
statement
is
true
or
false:
the
length
of
one
side
of
an
equilateral
triangle
determines
the
measurement
of
one
angle
in
hyperbolic
geometry.
Then,
the
longer
the
length
of
a
side
of
an
equilateral
triangle
the
larger
the
measurement
of
angle
is.
(Taki)
27. (Katy
K.)
MAT
211:
Final
Exam
Review
Student‐Written
Solutions
Fall
2010
Chapter
1
(Intro)
1. (Katy
Spencer)
2. (Maggie)
1.
Consider
the
parallelogram
ABCD
with
side
AB
parallel
to
side
DC
and
side
AD
parallel
to
side
BC
and
diagonals
AC
and
BD.
2.
By
the
alternate
interior
angle
theorem,
we
know
that
angle
ADM
is
congruent
to
angle
CBM
and
also,
angle
DAM
is
congruent
to
angle
BCM.
(Angle
ABM
congruent
to
angle
CDM.
Angle
DCM
is
congruent
to
angle
BAM).
3.
By
the
definition
of
parallel
lines,
we
know
that
Lines
are
parallel
if
they
lie
in
the
same
plane,
and
are
the
same
distance
apart
over
their
entire
length.
Hence,
we
know
that
AD
is
congruent
to
BC
and
AB
is
congruent
to
DC.
4.
By
ASA,
we
know
that
triangle
AMD
is
congruent
to
triangle
BMC
and
also
that
triangle
AMB
is
congruent
to
triangle
DMC.
By
CPCTC,
we
know
that
line
AM
is
congruent
to
line
MC
and
line
BM
is
congruent
to
line
MD.
5.
Therefore,
since
these
are
congruent,
we
know
that
the
diagonals
of
parallelogram
ABCD
bisect
each
other.
3. (Greg)
Finite
geometry
(Fe‐Fo)
4. (Ryan)
Conjecture
1:
No
one
line
contains
all
n
points.
Proof
by
contradiction:
Assume
one
line
does
contain
all
n
points.
Since
n
>
2,
this
contradicts
Axiom
3,
which
states,
“each
line
contains
exactly
two
points.”
Thus,
by
disproving
our
assumption,
we
prove
that,
“no
one
line
contains
all
n
points.”
Conjecture
2:
There
exist
exactly
[1
+
2
+
…
+
(n‐1)]
lines
in
the
system.
Conjecture
3:
Any
one
point
lies
on
exactly
(n‐1)
lines.
5. (Stephanie
M.)
Case
1:
Try
and
prove
that
for
every
group
of
three
athletes,
there
are
less
than
10
sports.
Case
2:
Try
and
prove
that
for
every
group
of
three
athletes,
there
are
more
than
10
sports.
Case
3:
There
are
exactly
ten
sports
they
have
shared.
Case
3:
The
athletes
are
Adam
(A),
Billy
(B),
Cody
(C),
Drew
(D),
and
Evan
(E).
This
supports
Axiom
1.
The
possible
groups
of
three
are
[ABC,
ABD,
ABE,
ACD,
ACE,
ADE,
BCD,
BCE,
BDE,
and
CDE].
This
supports
Axiom
2.
In
order
for
Axiom
3
to
be
correct,
there
must
be
ten
different
sports
because
not
all
of
the
athletes
have
played
the
same
sport.
6. (Stephanie
S.)
Theorem
1.
Assume
two
distinct
colleges
do
not
contain
exactly
one
student.
This
leads
to
two
cases.
Case
1‐
Two
distinct
colleges
contain
zero
student.
But
this
violates
axiom
4,
which
states
“Any
two
distinct
colleges
contain
at
least
one
student
which
belongs
to
both”
Case
2‐
Two
distinct
colleges
contain
more
than
one
student.
We
know
that
there
are
three
students
in
the
system
by
axiom
1.
So
we
only
need
to
consider
whether
two
distinct
colleges
contain
two
students
or
three
students.
But
by
axiom
3,
“Not
all
students
belong
to
the
same
college”
we
know
that
two
distinct
colleges
cannot
contain
three
students.
And
two
distinct
colleges
cannot
contain
two
students,
because
this
violates
axiom
2.
Thus,
because
both
cases
lead
to
contradictions
out
original
assumption
must
have
been
false.
So,
two
distinct
colleges
contain
exactly
one
student.
Theorem
2.
Assume
a
college
can
contain
three
distinct
students.
This
leads
to
two
cases,
because
by
axiom
1
we
know
that
there
are
exactly
three
students
and
three
colleges.
Case1‐
all
students
belong
to
one
college,
but
this
violates
axiom
3
“Not
all
students
belong
to
the
same
college”
Case
2‐
Each
college
contains
one
student,
this
violates
axiom
2
“Two
distinct
students
belong
to
exactly
one
college”
Thus,
since
both
cases
led
to
contradictions
our
original
assumption
must
have
been
false.
There
a
college
cannot
contain
three
distinct
students.
Theorem
3.
Assume
there
does
not
exist
a
set
of
two
colleges
that
contain
all
the
students
of
the
system.
This
leads
to
two
cases.
Case
1‐
one
college
contains
all
students,
this
violates
axiom
3
“Not
all
students
belong
to
the
same
college”
Case
2‐
Every
college
contains
just
one
student.
This
violates
axiom
2
“Two
distinct
students
belong
to
exactly
one
college”
and
we
know
there
are
only
three
students
because
of
axiom
1.
Thus,
because
the
two
cases
lead
to
contradictions
and
with
axiom
4
“Any
two
distinct
colleges
contain
at
least
one
student
which
belongs
to
both”
and
since
there
are
only
three
colleges
stated
by
axiom
1.
Out
original
assumption
is
false.
Therefore
there
exists
a
set
of
two
colleges
that
contain
all
the
students
of
the
system.
Chapter
2
(Triangles)
7. (Ellen)
Draw
triangle
ABC
with
incenter
I.
Drop
a
perpendicular
from
I
to
each
side
of
the
triangle.
Label
the
points
at
which
these
perpendiculars
and
sides
meet
as
M,
N,
and
P,
as
shown
in
the
figure.
We
seek
to
prove
that
MI
is
congruent
to
NI
which
is
congruent
to
PI,
because
a
perpendicular
is
the
shortest
distance
between
a
point
and
a
line
(as
shown
in
previous
homework)
and
therefore
if
MI,
NI,
and
PI
are
congruent,
I
is
equidistant
from
each
side.
Angle
AMI
and
angle
API
are
right
angles
because
they
are
formed
by
perpendiculars.
Angle
AMI
and
angle
API
are
congruent
because
all
right
angles
are
congruent,
according
to
Euclid’s
4th
postulate.
Angle
MAI
and
angle
PAI
are
congruent
because
they
are
formed
by
an
angle
bisector.
AI
is
an
angle
bisector
because
the
incenter
is
formed,
by
definition,
by
the
angle
bisectors
of
the
triangle.
AI
is
congruent
to
AI
by
the
reflexive
property.
Therefore,
by
AAS
congruent,
triangle
AMI
and
triangle
API
are
congruent.
By
CPCTC,
MI
is
congruent
to
PI.
Similarly,
angle
BMI
and
angle
BNI
are
congruent
because
they
are
right
angles,
by
Euclid’s
4th
postulate.
Angle
MBI
and
angle
NBI
are
congruent
because
they
are
formed
by
an
angle
bisector.
BI
is
congruent
to
BI
by
the
reflexive
property.
Therefore,
triangle
MBI
and
triangle
NBI
are
congruent
by
AAS
congruence.
By
CPCTC,
MI
is
congruent
to
NI.
Therefore,
since
MI
is
congruent
to
PI
and
MI
is
congruent
to
NI,
by
the
transitive
property,
MI,
NI,
and
PI
are
congruent
to
each
other
and
so
the
incenter
is
equidistant
from
every
side
of
the
triangle.
8. (Toni)
1.
True:
We
showed
in
class
that
the
diagonals
of
a
parallelogram
always
bisect
each
other
Implication
i. Hypothesis:
A
is
a
parallelogram
ii. Conclusion:
The
diagonals
bisect
each
other
iii. Converse:
If
the
diagonals
of
an
object
bisect
each
other
than
the
object
is
a
parallelogram
iv. Contrapositive:
If
the
diagonals
of
an
object
do
not
bisect
each
other
than
the
object
is
not
a
parallelogram
2.
False:
Not
all
right
triangles
are
isosceles,
meaning
that
they
can
have
two
angles
that
do
not
always
equal
each
other
Implication
v. Hypothesis:
A
is
a
right
triangle
vi. Conclusion:
A
is
an
isosceles
triangle
vii. Converse:
If
A
is
always
an
isosceles
triangle
than
A
is
a
right
triangle
viii. Contrapositive:
If
A
is
not
an
isosceles
triangle
than
A
is
not
a
right
triangle
3.
True:
We
showed
in
class
that
this
is
true
Not
an
implication
4.
False:
all
squares
are
rhombi
but
not
all
rhombi
are
squares
Not
an
implication
5.
False:
every
rectangle
does
have
three
side
BUT
all
right
triangles
do
not
have
three
angles
that
are
equal
(it
is
impossible
to
have
a
triangle
that
has
3‐
90
degree
angles)
Not
an
implication
9. (Anna)
We
have
triangle
ABC
and
triangle
DEF,
such
that
angle
A
is
congruent
to
angle
D,
side
AB
is
congruent
to
side
DE
and
angle
B
is
congruent
to
angle
E.
Assume
that
side
BC
is
not
congruent
to
side
EF.
Then
triangle
ABC
is
not
congruent
to
triangle
DEF,
as
congruent
triangles
have
corresponding
sides
and
angles
that
are
congruent.
There
exists
some
point
X
that
is
not
the
same
as
point
F
on
ray
EF
such
that
EX
is
congruent
to
BC.
X
must
either
be
between
E
and
F,
or
not
between
E
and
F
on
ray
EF.
Since
BC
is
congruent
to
EX,
angle
B
is
congruent
to
angle
E,
and
AB
is
congruent
to
DE,
triangle
ABC
is
congruent
to
triangle
DEX
by
SAS.
By
triangle
congruence,
angle
BAC
is
congruent
to
angle
EDX.
Case
1:
X
is
between
E
and
F
By
the
angle
addition
postulate,
angle
EDX
+
angle
XDF
equals
angle
EDF.
However,
since
angle
BAC
=
angle
EDF,
and
angle
EDX=angle
BAC,
we
have
that
angle
BAC
+
angle
XDF
equals
angle
BAC,
which
implies
that
angle
XDF
has
a
measure
of
of
zero
degrees,
but
this
is
not
the
case,
so
there
is
a
contradiction.
Case
2:
X
is
not
between
E
and
F,
but
on
ray
EF.
By
the
angle
addition
postulate,
angle
EDF
+
angle
FDX
equals
angle
EDX.
Since
angle
BAC
=
angle
EDF
and
angle
BAC=
angle
EDX,
angle
BAC
+
angle
FDX
equals
angle
BAC,
so
angle
FDX
has
a
measure
of
zero
degrees.
Since
F
is
not
the
same
point
as
X,
FDX
has
a
nonzero
measure,
so
this
is
a
contradiction.
Because
both
potential
cases
resulted
in
contradictions,
it
must
be
that
BC
is
congruent
to
EF.
Therefore,
by
SAS,
triangle
ABC
is
congruent
to
triangle
DEF.
This
means
that
if
two
triangles
have
the
conditions
of
ASA,
then
they
must
be
congruent.
10. (Ketty)
Chapter
3
(Circles)
11. (Clarissa)
12. (Mike)
13. (Caitlin)
Proof:
We
need
to
show
that
ray
XR
bisects
Angle
YXZ.
Drop
perpendiculars
from
R
to
lines
XY,
YZ,
XZ,
and
label
the
feet
of
these
perpendiculars
D,
E,
and
F
respectively.
Triangle
RDX
and
Triangle
RFX
are
right
triangles
by
construction.
RX=RX
by
the
reflexive
property.
Triangle
RDY=Triangle
REY
by
AAS.
Triangle
RFZ=Triangle
REZ,
also
by
AAS.
RD=RE,
and
RE=Rf,
since
they
are
corresponding
sides
of
congruent
triangles.
Thus,
RD=RF
by
the
transitive
property.
Therefore,
Triangle
RDX=Triangle
RFX
by
HL.
So
we
can
conclude
that
Angle
YXR=Angle
ZXR,
which
means
that
XR
bisects
Angle
YXZ.
14. (Joe
K.)
Chapter
4
(Analytic
geometry)
15. (Adam) Proof: Plot the points A(0,0), B(0,a), and C(b,0). Let the slope of segment
AB be called m1. Using the slope formula, m1= (a-0)/(0-0) = a/0, which is undefined.
So segment AB is vertical. Let the slope of segment AC be called m2. Using the
slope formula, m2 = (0-0)/(b-0) = 0/b = 0. So segment AC is horizontal. Since AB is
vertical and AC is horizontal, AB is perpendicular to AC and angle A is a right angle.
Therefore, triangle ABC is a right triangle. Using the distance formula,
,
, and
. So by substitution
and
Pythagorean theorem.
16. (Kevin)
. Therefore,
,
, which is the
17. (Katie
Skerik)
18. (Jamie)
a.)
Construct
a
right
triangle
ABC,
so
that
point
B
is
at
the
origin
(0,0).
Have
point
A
be
on
the
y‐axis
and
point
C
be
on
the
x‐axis,
so
that
they
have
the
coordinates
A(0,2a)
and
C(2c,0).
See
the
diagram
above
for
a
visual.
We
know
that
∠ABC
is
a
right
angle
because
it
is
formed
by
the
lines
BC
and
BA
and
:
The
slope
of
BC
is
(0‐0/0‐2c)
=
0
The
slope
of
BA
is
(0‐2a/0‐0)
=
Undefined
Since
lines
with
a
slope
of
zero
are
perpendicular
to
lines
with
an
undefined
slope,
we
know
BC
is
perpendicular
to
BA.
Therefore,
they
form
a
right
angle
at
their
intersection,
so
∠ABC
is
a
right
angle.
Therefore,
this
triangle
can
be
any
right
triangle,
as
long
as
A
is
on
the
y‐axis
and
C
is
on
the
x‐axis
and
B
is
the
only
point
at
the
origin.
b.)
Using
the
midpoint
formula,
we
can
find
the
midpoint
of
the
hypotenuse
AC:
((0+2c/2),
(2a
+
0/2)
)
=
(c,
a).
We
can
call
this
midpoint
point
M.
Chapter
6
(Transformational
geometry)
19. (Justin)
1.
Orientation:
reverse;
Fixed
point:
None
2.
Orientation:
reverse;
Fixed
point:
If
the
point
of
rotation
is
on
the
line
of
reflection
it
will
be
a
fixed
point.
3.
Orientation:
same;
Fixed
point:
There
is
a
point
in
the
plane
such
that
this
composition
could
have
been
just
a
single
rotation
about
a
point.
20. (Joe
H.)
A
translation‐translation
composition
can
be
replaced
by
a
translation.
A
reflection‐reflection
composition
over
interesting
lines
can
be
replaced
by
a
reflection.
A
reflection‐reflection
composition
over
parallel
lines
can
be
replaced
by
a
translation.
A
translation‐rotation
composition
can
be
replaced
by
a
rotation.
21. (Ryan
F.)
There are two cases in which a composition of two reflections will have
fixed points.
Case1) If the lines of reflection intersect, the point at which they intersect would be
fixed because we know that with a single reflection the points on the line of reflection
are fixed, then the common point between the two lines will also be fixed.
Case2) When the second reflection is reflected over the same line as the first
reflection, the only fixed points would be those contained on the line of reflection. It
is the same as saying that the two lines infinitely intersect and like in case1, where
they intersect will not be reflected, they will be fixed.
Chapter
9
(Hyperbolic
and
Spherical
Geometry)
22. (Kim)
1.
False
by
definition.
A
superparallel
is
any
ray
that
is
not
a
limiting
parallel
ray,
which
goes
through
a
point
P
and
is
parallel
to
l.
Because
a
limiting
parallel
is
a
ray
that
originates
at
point
P
and
is
the
closest
possible
ray
to
a
line
l
without
intersecting
it,
a
limiting
parallel
ray
would
be
the
answer
to
this
question,
not
superparallel.
2.
Sometimes.
This
statement
is
true
when
line
m
is
concurrent
with
line
k.
This
statement
is
false
when
line
m
is
not
concurrent
with
line
k.
This
is
because
in
hyperbolic
geometry
rectangles
do
not
exist.
Rectangles
do
not
exist
because
all
triangles
in
hyperbolic
have
a
positive
defect.
If
we
draw
line
segment
AC
we
can
see
this
as
two
triangles.
Since
triangles
must
have
a
positive
defect
their
angle
sum
is
always
less
than
180,
and
never
equal
to
180.
Since
defects
are
additive
we
can
see
this
also
as
adding
the
sums
of
the
two
triangles.
The
sum
of
this
quadrilateral
would
be
a+b+c+d+e+f.
We
know
that
angle
b
is
90˚
since
it
is
a
right
angle.
This
is
also
true
for
d
and
(c+e).
Thus
we
have
90+90+90+a+f.
We
also
know
that
these
two
triangles
together
must
be
less
than
360
(sum
of
two
triangles
that
are
less
than
180).
Thus
360>270+ + and
90> + .
Therefore
the
angle
DAB
cannot
be
a
right
angle
and
thus
there
is
not
a
second
common
parallel
between
line
j
and
l.
3.
The
defect
of
triangle
ABD
is
180‐a‐b‐f.
The
defect
of
triangle
BDC
is
180‐c‐d‐e.
The
defect
of
triangle
ABC
is
the
sum
of
the
defects
of
ABD
and
BCD.
Thus
we
have
the
defect
of
triangle
ABC
is
180‐a‐b‐f+180‐c‐d‐e.
From
here
we
can
see
that
the
angles
of
one
triangle
plus
the
angles
of
a
second
triangle
will
be
larger
than
either
of
the
individual
triangles.
Thus
the
angles
of
triangle
ABD
sum
to
a
smaller
number
than
the
angles
of
triangle
ABC.
Thus
if
a+b+f<a+b+c+d
then
the
defects
of
these
triangles
will
be
180‐a‐b‐f
for
triangle
ABD
and
180‐a‐b‐c‐d
for
triangle
ABC.
If
we
set
these
equations
equal
to
each
other
for
now
with
a
question
over
the
equals
sign
(in
this
case
a
‘does
not
equal’
sign)
we
get
180‐a‐b‐c‐d≠180‐a‐b‐f.
Doing
some
manipulation
we
get
a+b+c+d≠a+b+f.
By
looking
at
our
earlier
equation
we
can
replace
this
does
not
equal
sign
to
a
greater
than
sign
and
a+b+c+d>a+b+f.
Therefore,
by
going
backwards
we
can
see
that
180‐a‐b‐c‐
d<180‐a‐b‐f,
and
thus
the
defect
of
triangle
ABD
is
greater
than
the
defect
of
triangle
ABC.
23. (Kaitlyn)
In
the
diagram
above,
triangle
ADE
and
triangle
ABC
are
constructed
such
that
angle
EDA
is
congruent
to
angle
CBA
and
angle
BCA
is
congruent
to
angle
DEA.
Angle
A
is
congruent
to
angle
A
by
the
reflexive
property.
In
Hyperbolic
Geometry,
the
angle
defect
is
found
by
subtracting
the
sum
of
the
angles
from
180
degrees.
Using
this
formula,
the
defect
of
triangle
ABC
=
180
–
m<A
–
m<CBA
–
m<BCA.
Similarly,
the
defect
of
triangle
AED
=
180
–
m<A
–
m<EDA
–
m<DEA.
The
angle
defect
of
a
quadrilateral
is
360
degrees
minus
the
sum
of
the
angles,
which
means
that
the
defect
of
quadrilateral
BCED
=
360
–
m<CBA
–
m<BCA
–
m<CED
–
m<BDE.
Angle
defects
are
additive,
which
means
that
the
defect
of
triangle
ABC
is
equal
to
the
sum
of
the
defects
of
triangle
ADE
and
quadrilateral
BCED.
Thus,
defect
of
triangle
ABC
=
180
–
m<A
–
m<EDA
–
m<DEA
+
360
–
m<CBA
–
m<BCA
–
m<CED
–
m<BDE.
After
combining
like
terms,
the
defect
of
triangle
ABC
=
540
–
m<A
–
(m<DEA
+
m<CED)
–
(m<EDA
+
m<BDE)
–
m<CBA
–
m<BCA.
Since
<AEC
and
<ADB
are
a
straight
angles,
m<AEC
=
180
degrees
=
m<ADB.
<AEC
and
<ADB
are
split
into
two
angles
by
segment
ED,
which
means
that
m<AEC
=
m<DEA
+
m<CED
and
m<ADB
=
m<EDA
+
m<BDE.
By
the
transitive
property,
m<DEA
+
m<CED
=
180
degrees
=
m<EDA
+
m<BDE.
By
substitution,
the
defect
of
triangle
ABC
=
540
–
m<A
–
(180)
–
(180)
–
m<CBA
–
m<BCA
=
180
–
m<A
–
m<CBA
–
m<BCA,
which
shows
that
the
defect
of
triangle
ADE
+the
defect
of
quadrilateral
BCED
=
the
defect
of
triangle
ABC.
This
means
that
the
defect
of
triangle
ABC
is
larger
than
the
defect
of
triangle
ADE,
which
in
turn
means
that
the
sum
of
the
angles
of
triangle
ABC
must
be
smaller
than
the
sum
of
the
angles
of
triangle
ADE.
This
shows
that
a
larger
triangle
cannot
have
the
same
angle
measure
as
a
smaller
triangle.
Similarly,
a
smaller
triangle
has
a
larger
angle
measure
than
a
larger
triangle.
Thus,
the
only
way
that
two
triangles
can
have
the
same
angle
measure
is
when
the
sides
are
also
congruent.
24. (Lina)
Construct
a
perpendicular
from
E
to
segment
CD
with
intersection
point,
R.
We
know
the
measure
of
angle
GRE
is
90
degrees
by
construction.
We
also
know
that
measure
of
angle
FRE
is
90
degrees
by
construction.
Since
the
base
angles
of
a
Saccheri
Quadrilateral
are
right
angles,
we
know
angle
BDG
is
congruent
to
angle
ERG
since
all
right
angles
are
congruent.
We
know
BG
is
congruent
to
EG
by
construction.
We
know
angle
RGE
is
congruent
to
angle
DGB
since
vertical
angles
are
congruent;
thus,
by
HA
triangle
GRE
is
congruent
to
triangle
GDB.
By
CPCTC,
angle
REG
is
congruent
to
angle
GBD.
Similarly,
we
know
that
angle
ACF
is
congruent
to
angle
FRE
since
all
right
angles
are
congruent.
We
know
AF
is
congruent
to
EF
by
construction.
Angle
AFC
is
congruent
to
angle
EFR
since
vertical
angles
are
congruent.
By
HA
triangle
ACF
is
congruent
to
triangle
ERF.
By
CPCTC,
angle
CAF
is
congruent
to
angle
FEF.
The
sum
of
the
summit
angles
is
equal
to
the
m<CAF
+
m<EAB
+
m<ABE
+
m<EBD,
by
substitution,
the
sum
is
equal
to
m<FER
+
m<EAB+
m<ABE
+
m<BER
which
is
the
sum
of
measures
of
the
angles
of
the
associated
triangle.
25. (Dan)
26. (Taki)
The
answer
is
false.
The
figure
is
a
counter
example
of
the
statement.
27. (Katy
K.)