Find each measure.
1.XW
SOLUTION:
Given that
By the Perpendicular Bisector Theorem, XW = XY.
Therefore, XW = 12.
ANSWER:
12
2.AC
SOLUTION:
In the figure, AB = BC. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, AC = DC.
Since AD = 7, DC = 7.
By the Segment Addition Postulate,
ANSWER:
14
3.LP
SOLUTION:
Given that
So LM = LP, by the Perpendicular Bisector
Theorem.
Therefore,
Solve for x.
Substitute x = 2 in the expression for LP.
ANSWER:
15
4.ADVERTISING Four friends are passing out flyers
at a mall food court. Three of them take as many
flyers as they can and position themselves as shown.
The fourth one keeps the supply of additional flyers.
Copy the positions of points A, B, and C. Then
position the fourth friend at D so that she is the same
distance from each of the other three friends.
SOLUTION:
You will need to find the circumcenter of the triangle
formed by points A, B, and C. This can be done by
constructing the perpendicular bisectors of each side
of the triangle and finding their point of concurrency.
Point D, as shown in the diagram, is where the 4th
friend should position herself so that she is equidistant
totheothers.
ANSWER:
Find each measure.
5.CP
SOLUTION:
Given: istheanglebisectorof ,
.
By the Angle Bisector Theorem,
So, CP = 8.
ANSWER:
8
6.
SOLUTION:
Given:
bisects bytheconverseoftheAngle
Bisector Theorem.
Therefore,
ANSWER:
7.QM
SOLUTION:
From the figure, istheanglebisectorof
.
By the Angle Bisector Theorem,
ANSWER:
12
8.CCSS SENSE-MAKING Find JQ if Q is the
incenter of .
SOLUTION:
Since Q is the incenter of ,
Use the Pythagorean Theorem in triangle JPQ.
ANSWER:
18.8
Find each measure.
9.NP
SOLUTION:
From the figure,
By the Perpendicular Bisector Theorem, LP = NP.
Substitute x = 9 in the expression for NP.
ANSWER:
14
10.PS
SOLUTION:
Given that
By the Perpendicular Bisector Theorem, PS = PQ.
Therefore, PS = 9.
ANSWER:
9
11.KL
SOLUTION:
Here JM = LM. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, JK = KL.
ANSWER:
6
12.EG
SOLUTION:
Given: GD = ED. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, GF = FE = 5.
ANSWER:
10
13.CD
SOLUTION:
Given: AB = AD. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, BC = CD
ANSWER:
4
14.SW
SOLUTION:
WT = TS by the converse of the Perpendicular
Bisector Theorem.
That is,
Solve for x.
Substitute x = 3 in the expression for WT.
ANSWER:
16
15.STATE FAIR The state fair has set up the location
of the midway, livestock competition, and food
vendors. The fair planners decide that they want to
locate the portable restrooms the same distance from
each location. Copy the positions of points M, L, and
F. Then find the location for the restrooms and label
it R.
SOLUTION:
You can find point R by constructing the
perpendicular bisectors of each side of the triangle
formed by points F, L, and M. Their point of
concurrency, R, is equidistant from each vertex of
thetriangle.
ANSWER:
16.SCHOOL A school system has built an elementary,
middle, and high school at the locations shown in the
diagram. Copy the positions of points E, M, and H.
Then find the location for the bus yard B that will
service these schools so that it is the same distance
from each school.
SOLUTION:
To find the best place for the bus yard, you will need
to construct the perpendicular bisector of each side
of the triangle. Their point of concurrency is point B,
which is the same distance from each point H, E, and
M.
ANSWER:
Point D is the circumcenter of . List any
segment(s) congruent to each segment.
17.
SOLUTION:
Circumcenter D is equidistant from the vertices of
the triangle ABC.
ANSWER:
18.
SOLUTION:
is the perpendicular bisector of . So,
ANSWER:
19.
SOLUTION:
is the perpendicular bisector of . So,
ANSWER:
20.
SOLUTION:
Circumcenter D is equidistant from the vertices of
the triangle ABC.
Therefore,
ANSWER:
Find each measure.
21.AF
SOLUTION:
By the Angle Bisector Theorem, AF = AD = 11.
ANSWER:
11
22.mDBA
SOLUTION:
by the converse of the
Angle Bisector Theorem.
ANSWER:
23.
SOLUTION:
The converse of the Angle Bisector Theorem says
That is,
Solve the equation for x.
Now,
ANSWER:
24.XA
SOLUTION:
By the Angle Bisector Theorem, XA = ZA= 4.
ANSWER:
4
25.mPQS
SOLUTION:
In triangle QRS,
Substitute the known values.
By the converse of the Angle Bisector Theorem,
Therefore
ANSWER:
26.PN
SOLUTION:
Here , by the Angle Bisector Theorem.
Substitute x = 8 in the expression for PN.
ANSWER:
30
CCSS SENSE-MAKINGPoint P is the incenter
of . Find each measure below.
27.PB
SOLUTION:
Use the Pythagorean Theorem in the right triangle
ABP.
ANSWER:
7.1
28.DE
SOLUTION:
is the angle bisector of BytheAngle
Bisector Theorem, PD = PB. Or PD = 7.1.
In triangle PDE,
ANSWER:
13.1
29.m DAC
SOLUTION:
By the Converse of the Angle Bisector Theorem,
istheanglebisectorof . Therefore,
m
ANSWER:
30. DEP
SOLUTION:
By SAS postulate,
So,
ANSWER:
31.INTERIOR DESIGN You want to place a
centerpiece on a corner table so that it is located the
same distance from each edge of the table. Make a
sketch to show where you should place the
centerpiece. Explain your reasoning.
SOLUTION:
Find the point of concurrency of the angle bisectors
of the triangle, the incenter. This point is equidistant
from each side of the triangle.
ANSWER:
Find the point of concurrency of the angle bisectors
of the triangle, the incenter. This point is equidistant
from each side of the triangle.
Determine whether there is enough information
given in each diagram to find the value of x.
Explain your reasoning.
32.
SOLUTION:
There is not enough information to find the value of
x. In order to determine the value of x, we would
need to see markings indicating whether the
segments drawn from the bisecting ray are
perpendicular to each side of the angle being
bisected. Since there are no markings indicating
perpendicularity,noconclusioncanbemade.
ANSWER:
No; we need to know if the segments are
perpendicular to the rays.
33.
SOLUTION:
There is not enough information to find the value of
x. In order to determine the value of x, we would
need to see markings indicating whether the
segments drawn from the ray are congruent to each
other. Since there are no markings indicating
congruence,noconclusioncanbemade.
ANSWER:
No; we need to know whether the perpendicular
segments are congruent to each other.
34.
SOLUTION:
There is not enough information to find the value of
x. In order to determine the value of x, we would
need to see markings indicating whether the segment
bisector is perpendicular to the side of the triangle
that is bisected. Since there are no markings
indicating perpendicularity, no conclusion can be
made.
ANSWER:
No; we need to know if the segment bisector is a
perpendicular bisector.
35.
SOLUTION:
There is not enough information to find the value of
x. According to the Converse of the Perpendicular
Bisector Theorem, we would need to know if the
point of the triangle is equidistant to the endpoints of
the segment. Since we don't know if the hypotenuses
of the two smaller triangles are congruent, we can't
assume that the other side of the big triangle is
bisected.
ANSWER:
No; we need to know whether the hypotenuses of
the triangles are congruent.
36.SOCCER A soccer player P is approaching the
opposing team’s goal as shown in the diagram. To
make the goal, the player must kick the ball between
the goal posts at L and R. The goalkeeper faces the
kicker. He then tries to stand so that if he needs to
dive to stop a shot, he is as far from the left-hand
side of the shot angle as the right-hand side.
a. Describe where the goalkeeper should stand.
Explain your reasoning.
b. Copy . Use a compass and a straightedge
to locate point G, the desired place for the
goalkeeper to stand.
c. If the ball is kicked so it follows the path from P to
R, construct the shortest path the goalkeeper should
take to block the shot. Explain your reasoning.
SOLUTION:
a. The goalkeeper should stand along the angle
bisector of the opponent’s shot angle, since the
distance to either side of the angle is the same along
this line.
b. To determine a point, construct the angle bisector
of ∠P and then mark point Ganywhereon .
c. We can find the shortest distance from point G to
either side of ∠RPL by constructing a line that is
perpendicular to each side of the angle that passes
through point G. The shortest distance to a line from
a point not on the line is the length of the segment
perpendicular to the line from the point
.
ANSWER:
a. The goalkeeper should stand along the angle
bisector of the opponent’s shot angle, since the
distance to either side of the angle is the same along
this line.
b.
c.
The shortest distance to a line from a point not on the
line is the length of the segment perpendicular to the
line from the point.
PROOF Write a two-column proof.
37.Theorem 5.2
Given:
Prove: C and D are on the perpendicular bisector
of
SOLUTION:
As in all proofs, you need to think backwards. What
would you need to do to prove that a segment in a
perpendicular bisector? You can prove it in two parts
- first, that and are perpendicular to each
other and then, that isbisected.Thiswillinvolve
proving two triangles are congruent so that you can
get congruent corresponding parts (CPCTC). Start
by considering which triangles you can make
congruent to each other using the given information.
There are three pairs of triangles in the diagram, so
mark the given information to lead you to the correct
pair. Once you have determined that
, then think about what congruent
corresponding parts you need to make
. To prove that two segments are
perpendicular, consider what CPCTC you need to
choose that would eventually prove that two
segments form a right angle. In addition, which
CPCTC would you choose to prove that a segment is
bisected?
Proof:
Statement(Reasons)
1. (Given)
2. (Congruence of segments is reflexive.)
3. (SSS)
4. (CPCTC)
5. (Congruence of segments is reflexive.)
6. (SAS)
7. (CPCTC)
8. E is the midpoint of . (Definition of midpoint)
9. (CPCTC)
10. formalinearpair.(Definition
of linear pair)
11. aresupplementary.
(Supplementary Theorem)
12. (Definition of
supplementary)
13. (Substitution
Property)
14. (Substitution Property)
15. (Division Property)
16. arerightangles.(Definitionof
right angle)
17. (Definition of Perpendicular lines)
18. istheperpendicularbisectorof .
(Definition of perpendicular bisector)
19. C and D are on the perpendicular bisector of
. (Definition of point on a line)
ANSWER:
Statement(Reasons)
1. (Given)
2. (Congruence of segments is reflexive.)
3. (SSS)
4. (CPCTC)
5. (Congruence of segments is reflexive.)
6. (SAS)
7. (CPCTC)
8. E is the midpoint of . (Definition of midpoint)
9. (CPCTC)
10. formalinearpair.(Definition
of linear pair)
11. aresupplementary.
(Supplementary Theorem)
12. (Definition of
supplementary)
13. (Substitution
Property)
14. (Substitution Property)
15. (DivisionProperty)
16. arerightangles.(Definitionof
right angle)
17. (DefinitionofPerpendicularlines)
18. istheperpendicularbisectorof .
(Definition of perpendicular bisector)
19. C and D are on the perpendicular bisector of
. (Definition of point on a line)
38.Theorem 5.6
Given: , angle bisectors and
Prove:
SOLUTION:
Consider what it means if are
allanglebisectorsandmeetatpointK.Thekeyto
thisproofistheAngleBisectorTheorem.
Proof:
Statement(Reasons)
1. anglebisectors and
(Given)
2. KP = KQ, KQ = KR, KP = KR (Any point on the
angle bisector is equidistant from the sides of the
angle.)
3. KP = KQ = KR(Transitive Property)
ANSWER:
Proof:
Statement(Reasons)
1. anglebisectors and
(Given)
2. KP = KQ, KQ = KR, KP = KR (Any point on the
angle bisector is equidistant from the sides of the
angle.)
3. KP = KQ = KR(Transitive Property)
CCSSARGUMENTSWriteaparagraphproof
of each theorem.
39.Theorem 5.1
SOLUTION:
If we know that is the perpendicular bisector to
, then what two congruent triangle pairs can we
get from this given information? Think about what
triangles we can make congruent to prove that
?
Given: isthe bisectorof
E is a point on CD.
Prove: EA = EB
Proof: isthe ⊥ bisector of Bydefinitionof
bisector, D is the midpoint of Thus,
bytheMidpointTheorem. CDA and
CDB are right angles by the definition of
perpendicular. Since all right angles are congruent,
CDA CDB. Since E is a point on
EDA and EDB are right angles and are
congruent. By the Reflexive Property,
Thus EDA EDB by SAS.
becauseCPCTC,andbydefinitionof
congruence, EA = EB.
ANSWER:
Given: isthe bisectorof
E is a point on CD.
Prove: EA = EB
Proof: isthe ⊥ bisector of Bydefinitionof
bisector, D is the midpoint of Thus,
bytheMidpointTheorem. CDA and
CDB are right angles by the definition of
perpendicular. Since all right angles are congruent,
CDA CDB. Since E is a point on
EDA and EDB are right angles and are
congruent. By the Reflexive Property,
Thus EDA EDB by SAS.
becauseCPCTC,andbydefinitionof
congruence, EA = EB.
40.Theorem 5.5
SOLUTION:
If PD and PE are equidistant to the sides of ,
then they are also perpendicular to those sides.If you
can prove , then consider what
CPCTC would need to be chosen to prove that
is an angle bisector of
Given: BAC
P is in the interior of BAC
PD = PE
Prove: istheanglebisectorof BAC
Proof: Point P is on the interior of BAC of
andPD = PE. By definition of congruence,
and sincethe
distance from a point to a line is measured along the
perpendicular segment from the point to the line.
ADP and AEP are right angles by the definition
of perpendicular lines and and are
right triangles by the definition of right triangles. By
the Reflexive Property, Thus,
byHL. DAP EAP because
CPCTC, and istheanglebisectorof BAC by
the definition of angle bisector.
ANSWER:
Given: BAC
P is in the interior of BAC
PD = PE
Prove: istheanglebisectorof BAC
Proof: Point P is on the interior of BAC of
andPD = PE. By definition of congruence,
. and sincethe
distance from a point to a line is measured along the
perpendicular segment from the point to the line.
ADP and AEP are right angles by the definition
of perpendicular lines and and are
right triangles by the definition of right triangles. By
the Reflexive Property, Thus,
byHL. DAP EAP because
CPCTC, and istheanglebisectorof BAC by
the definition of angle bisector.
COORDINATE GEOMETRY Write an
equation in slope-intercept form for the
perpendicular bisector of the segment with the
given endpoints. Justify your answer.
41.A(–3, 1) and B(4, 3)
SOLUTION:
The slope of the segment AB is or So,the
slope of the perpendicular bisector is
The perpendicular bisector passes through the
midpoint of the segment AB. The midpoint of AB is
or
The slope-intercept form for the equation of the
perpendicular bisector of AB is:
ANSWER:
; The perpendicular bisector bisects
the segment at the midpoint of the segment. The
midpoint is . The slope of the given segment
is , so the slope of the perpendicular bisector is
.
42.C(–4, 5) and D(2, –2)
SOLUTION:
The slope of the segment CD is or So,
the slope of the perpendicular bisector is
The perpendicular bisector passes through the
midpoint of the segment CD. The midpoint of CD is
or
The slope-intercept form for the equation of the
perpendicular bisector of CD is:
ANSWER:
; The perpendicular bisector bisects the
segment at the midpoint of the segment. The
midpoint is .Theslopeofthegivensegment
is ,sotheslopeoftheperpendicularbisector
is .
43.PROOF Write a two-column proof of Theorem 5.4.
SOLUTION:
The key to this proof is to think about how to get
. Use your given information to
get congruent corresponding parts. Remember that
an angle bisector makes two anglescongruent.
Given: bisects QPR. and
Prove:
Proof:
Statements (Reasons)
1. bisects QPR,and .
(Given)
2. (Definition of angle bisector)
3. PYX and PZX are right angles. (Definition of
perpendicular)
4. (Right angles are congruent.)
5. (Reflexive Property)
6.
7. (CPCTC)
ANSWER:
Given: bisects QPR. and
.
Prove:
Proof:
Statements (Reasons)
1. bisects QPR,and .
(Given)
2. (Definition of angle bisector)
3. PYX and PZX are right angles. (Definition of
perpendicular)
4. (Right angles are congruent.)
5. (Reflexive Property)
6.
7. (CPCTC)
44.GRAPHIC DESIGN Mykia is designing a pennant
for her school. She wants to put a picture of the
school mascot inside a circle on the pennant. Copy
the outline of the pennant and locate the point where
the center of the circle should be to create the largest
circle possible. Justify your drawing.
SOLUTION:
We need to find the incenter of the triangle by finding
the intersection point of the angle bisectors. To make
the circle, place your compass on the incenter, mark
a radius that is perpendicular to each side (the
shortest length) and make a circle.When the circle is
as large as possible, it will touch all three sides of the
pennant.
ANSWER:
When the circle is as large as possible, it will touch
all three sides of the pennant. We need to find the
incenter of the triangle by finding the intersection
point of the angle bisectors.
COORDINATE GEOMETRY Find the
coordinates of the circumcenter of the triangle
with the given vertices. Explain.
45.A(0, 0), B(0, 6), C(10, 0)
SOLUTION:
Graph the points.
Since the circumcenter is formed by the
perpendicular bisectors of each side of the triangle,
we need to draw the perpendicular bisectors of the
two legs of the triangle and see where they intersect.
The equation of a line of one of the perpendicular
bisectors is y = 3 because it is a horizontal line
throughpoint(0,3),themidpointoftheverticalleg.
The equation of a line of another perpendicular
bisector is x = 5, because it is a vertical line through
thepoint(5,0),themidpointofthehorizontalleg.
These lines intersect at (5, 3), because it will have
the x-value of the line x=5 and the y-value of the line
y=3.
The circumcenter is located at (5, 3).
ANSWER:
The equation of a line of one of the perpendicular
bisectors is y = 3. The equation of a line of another
perpendicular bisector is x = 5. These lines intersect
at (5, 3). The circumcenter is located at (5, 3).
46.J(5, 0), K(5, –8), L(0, 0)
SOLUTION:
Since the circumcenter is formed by the
perpendicular bisectors of each side of the triangle,
we need to draw the perpendicular bisectors of the
two legs of the triangle and see where they intersect.
The equation of a line of one of the perpendicular
bisectors is y = –4 because it is a horizontal line
through point (5, –4),themidpointoftheverticalleg.
The equation of a line of another perpendicular
bisector is x = 2.5, because it is a vertical line through
thepoint(2.5,0),themidpointofthehorizontalleg.
These lines intersect at (2.5, –4), because it will have
the x-value of the line x = 2.5 and the y-value of the
line y = –4.
The circumcenter is located at (2.5, –4).
ANSWER:
The equation of a line of one perpendicular bisector
is y = –4. The equation of a line of another
perpendicular bisector is x = 2.5. These lines
intersect at (2.5, –4). The circumcenter is located at
(2.5, –4).
47.LOCUS Consider .Describethesetofall
points in space that are equidistant from C and D.
SOLUTION:
Since we are considering all points in space, we
need to consider more than just the perpendicular
bisector of CD. The solution is a plane that is
perpendicular to the plane in which liesand
bisects .
ANSWER:
a plane perpendicular to the plane in which lies
and bisecting .
48.ERROR ANALYSIS Claudio says that from the
information supplied in the diagram, he can conclude
that K is on the perpendicular bisector of .
Caitlyn disagrees. Is either of them correct? Explain
your reasoning.
SOLUTION:
Caitlyn is correct; Based on the markings, we only
know that J is the midpoint of . We don't know if
is perpendicular to .
ANSWER:
Caitlyn; K is only on the perpendicular bisector of
if , but we are not given this
information in the diagram.
49.OPEN ENDED Draw a triangle with an incenter
located inside the triangle but a circumcenter located
outside. Justify your drawing by using a straightedge
and a compass to find both points of concurrency.
SOLUTION:
Knowing that the incenter of a triangle is always
found inside any triangle, the key to this problem is to
think about for which type of triangle the
circumcenter would fall on the outside. Experiment
with acute, right and obtuse triangles to find the one
thatwillwork.
ANSWER:
Sample answer:
CCSS ARGUMENTS Determine whether each
statement is sometimes, always, or never true.
Justify your reasoning using a counterexample
or proof.
50.The angle bisectors of a triangle intersect at a point
that is equidistant from the vertices of the triangle.
SOLUTION:
It is the circumcenter, formed by the perpendicular
bisectors, that is equidistant to the vertices of a
triangle. Consider for what type of triangle the
incenter, formed by the angle bisectors, would be the
same location as the circumcenter. Sometimes; if the
triangle is equilateral, then this is true, but if the
triangle is isosceles or scalene, the statement is false.
.
JQ = KQ = LQ
ANSWER:
Sometimes; if the triangle is equilateral, then this is
true, but if the triangle is isosceles or scalene, the
statement is false.
.
JQ = KQ = LQ
51.In an isosceles triangle, the perpendicular bisector of
the base is also the angle bisector of the opposite
vertex.
SOLUTION:
Always. It is best to show this through a proof that
would show that, given an isosceles triangle divided
by a perpendicular bisector, that it creates two
congruent triangles. Therefore, by CPCTC, the two
angles formed by the perpendicular bisector are
congruentandformananglebisector.
Given: is isosceles with legs and ;
is the bisector of .
Prove: is the angle bisector of ABC.
Proof:
Statements (Reasons)
1. isisosceleswithlegs and . (Given)
2. (Definition of isosceles triangle)
3. is the bisector of .(Given)
4. D is the midpoint of . (Definition of segment
bisector)
5. (Definition of midpoint)
6. (Reflexive Property)
7.
8.
9. is the angle bisector of ABC. (Definition of
angle bisector)
ANSWER:
Always.
Given: is isosceles with legs and ;
is the bisector of .
Prove: is the angle bisector of ABC.
Proof:
Statements (Reasons)
1. isisosceleswithlegs and . (Given)
2. (Def. of isosceles )
3. is the bisector of .(Given)
4. D is the midpoint of . (Def. of segment
bisector)
5. (Def. of midpoint)
6. (Reflexive Property)
7.
8.
9. is the angle bisector of ABC. (Def.
bisector)
CHALLENGE Write a two-column proof for
each of the following.
52.Given: Plane Yisaperpendicular
bisector of .
Prove:
SOLUTION:
A plane that is a perpendicular bisector of a segment
behaves the same way as a segment that is a
perpendicular bisector. It creates two right angles
and two congruent corresponding segments. Use
them to prove two triangles congruent in this
diagram.
Proof:
Statements (Reasons)
1. Plane Y is a perpendicular bisector of . (Given)
2. DBA and CBA are right angles,
(Definitionof bisector)
3. (Right angles are congruent.)
4. (Reflexive Property)
5. (SAS)
6. (CPCTC)
ANSWER:
Proof:
Statements (Reasons)
1. Plane Y is a perpendicular bisector of . (Given)
2. DBA and CBA are rt. s, (Def.
of bisector)
3. (Right angles are congruent.)
4. (Reflexive Property)
5. (SAS)
6. (CPCTC)
53.Given: Plane Z is an angle bisector of
.
Prove:
SOLUTION:
A plane that is an angle bisector of a segment
behaves the same way as a segment that is an angle
bisector. It creates two congruent angles. Use these
angles, along with the other given and the diagram, to
prove two triangles congruent.
Proof:
Statements (Reasons)
1. Plane Z is an angle bisector of KJH;
(Given)
2. (Definition of angle bisector)
3. (Reflexive Property)
4.
5. (CPCTC)
ANSWER:
Proof:
Statements (Reasons)
1. Plane Z is an angle bisector of KJH;
(Given)
2. (Definitionofanglebisector)
3. (ReflexiveProperty)
4.
5. (CPCTC)
54.WRITING IN MATH Compare and contrast the
perpendicular bisectors and angle bisectors of a
triangle. How are they alike? How are they
different? Be sure to compare their points of
concurrency.
SOLUTION:
Start by making a list of what the different bisectors
do. Consider what they bisect, as well as the point of
concurrency they form. Discuss the behaviors of
these points of concurrency in different types of
triangles. Are they ever found outside the triangle?
Insidethetriangle?Onthetriangle?
The bisectors each bisect something, but the
perpendicular bisectors bisect segments while angle
bisectors bisect angles. They each will intersect at a
point of concurrency. The point of concurrency for
perpendicular bisectors is the circumcenter. The
point of concurrency for angle bisectors is the
incenter. The incenter always lies in the triangle,
while the circumcenter can be inside, outside, or on
the triangle.
ANSWER:
The bisectors each bisect something, but the
perpendicular bisectors bisect segments while angle
bisectors bisect angles. They each will intersect at a
point of concurrency. The point of concurrency for
perpendicular bisectors is the circumcenter. The
point of concurrency for angle bisectors is the
incenter. The incenter always lies in the triangle,
while the circumcenter can be inside, outside, or on
the triangle.
55.ALGEBRA An object is projected straight upward
with initial velocity v meters per second from an
initial height of s meters. The distance d in meters
the object travels after t seconds is given by d = –
10t2+ vt + s. Sherise is standing at the edge of a
balcony 54 meters above the ground and throws a
ball straight up with an initial velocity of 12 meters
per second. After how many seconds will it hit the
ground?
A 3 seconds
B 4 seconds
C 6 seconds
D 9 seconds
SOLUTION:
Substitute the known values in the distance
expression.
Solve the equation for t.
Discard the negative value, as it has no real world
meaning in this case.
The correct choice is A.
ANSWER:
A
56.SAT/ACT For
F x + 12
G x + 9
H x + 3
J x
K 3
SOLUTION:
The correct choice is K.
ANSWER:
K
57.A line drawn through which of the following points
would be a perpendicular bisector of ?
A T and K
B L and Q
C J and R
D S and K
SOLUTION:
The line passes through the points K and S will be a
perpendicular bisector of
The correct choice is D.
ANSWER:
D
58.SAT/ACT For x≠–3,
Fx + 12
Gx + 9
Hx + 3
J x
K3
SOLUTION:
Simplify the expression by first removing the GCF.
Therefore, the correct choice is K.
ANSWER:
K
Name the missing coordinate(s) of each
triangle.
59.
SOLUTION:
The line drawn through the point L is the
perpendicular bisector of Therefore,x-
coordinate of L is x-coordinate of the midpoint of JK.
The x-coordinate of L is ora.
Thus, the coordinates of L is (a, b).
ANSWER:
L(a, b)
60.
SOLUTION:
Since AC = AB, the distance from the origin to point
B and C is the same. They are both a length away
from the origin. The coordinate of C is (a, 0) because
it is on the x-axis and is a units away from the
origin.
ANSWER:
C(a, 0)
61.
SOLUTION:
Since the triangle is an isosceles triangle, S and T are
equidistant from O. So, the coordinates of S are (–
2b, 0). R lies on the y-axis. So, its x-coordinate is 0.
The coordinates of R are (0, c)
ANSWER:
S(–2b, 0) and R(0, c)
COORDINATE GEOMETRY Graph each pair
of triangles with the given vertices. Then
identify the transformation and verify that it is a
congruence transformation.
62.A(–2, 4), B(–2, –2), C(4, 1);
R(12, 4), S(12, –2), T(6, 1)
SOLUTION:
is a reflection of .
Findthelengthofeachside.
by SSS.
ANSWER:
is a reflection of ; AB = 6, BC = ,
AC = , TR = , RS = 6, TS= .
by SSS.
63.J(–3, 3), K(–3, 1), L(1, 1);
X(–3, –1), Y(–3, –3), Z(1, –3)
SOLUTION:
is a translation of .
Find the length of each side.
by SSS.
Find each measure.
1.XW
SOLUTION:
Given that
By the Perpendicular Bisector Theorem, XW = XY.
Therefore, XW = 12.
ANSWER:
12
2.AC
SOLUTION:
In the figure, AB = BC. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, AC = DC.
Since AD = 7, DC = 7.
By the Segment Addition Postulate,
ANSWER:
14
3.LP
SOLUTION:
Given that
So LM = LP, by the Perpendicular Bisector
Theorem.
Therefore,
Solve for x.
Substitute x = 2 in the expression for LP.
ANSWER:
15
4.ADVERTISING Four friends are passing out flyers
at a mall food court. Three of them take as many
flyers as they can and position themselves as shown.
The fourth one keeps the supply of additional flyers.
Copy the positions of points A, B, and C. Then
position the fourth friend at D so that she is the same
distance from each of the other three friends.
SOLUTION:
You will need to find the circumcenter of the triangle
formed by points A, B, and C. This can be done by
constructing the perpendicular bisectors of each side
of the triangle and finding their point of concurrency.
Point D, as shown in the diagram, is where the 4th
friend should position herself so that she is equidistant
totheothers.
ANSWER:
Find each measure.
5.CP
SOLUTION:
Given: istheanglebisectorof ,
.
By the Angle Bisector Theorem,
So, CP = 8.
ANSWER:
8
6.
SOLUTION:
Given:
bisects bytheconverseoftheAngle
Bisector Theorem.
Therefore,
ANSWER:
7.QM
SOLUTION:
From the figure, istheanglebisectorof
.
By the Angle Bisector Theorem,
ANSWER:
12
8.CCSS SENSE-MAKING Find JQ if Q is the
incenter of .
SOLUTION:
Since Q is the incenter of ,
Use the Pythagorean Theorem in triangle JPQ.
ANSWER:
18.8
Find each measure.
9.NP
SOLUTION:
From the figure,
By the Perpendicular Bisector Theorem, LP = NP.
Substitute x = 9 in the expression for NP.
ANSWER:
14
10.PS
SOLUTION:
Given that
By the Perpendicular Bisector Theorem, PS = PQ.
Therefore, PS = 9.
ANSWER:
9
11.KL
SOLUTION:
Here JM = LM. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, JK = KL.
ANSWER:
6
12.EG
SOLUTION:
Given: GD = ED. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, GF = FE = 5.
ANSWER:
10
13.CD
SOLUTION:
Given: AB = AD. By the converse of the
Perpendicular Bisector Theorem, isa
perpendicular bisector of
Therefore, BC = CD
ANSWER:
4
14.SW
SOLUTION:
WT = TS by the converse of the Perpendicular
Bisector Theorem.
That is,
Solve for x.
Substitute x = 3 in the expression for WT.
ANSWER:
16
15.STATE FAIR The state fair has set up the location
of the midway, livestock competition, and food
vendors. The fair planners decide that they want to
locate the portable restrooms the same distance from
each location. Copy the positions of points M, L, and
F. Then find the location for the restrooms and label
it R.
SOLUTION:
You can find point R by constructing the
perpendicular bisectors of each side of the triangle
formed by points F, L, and M. Their point of
concurrency, R, is equidistant from each vertex of
thetriangle.
ANSWER:
16.SCHOOL A school system has built an elementary,
middle, and high school at the locations shown in the
diagram. Copy the positions of points E, M, and H.
Then find the location for the bus yard B that will
service these schools so that it is the same distance
from each school.
SOLUTION:
To find the best place for the bus yard, you will need
to construct the perpendicular bisector of each side
of the triangle. Their point of concurrency is point B,
which is the same distance from each point H, E, and
M.
ANSWER:
Point D is the circumcenter of . List any
segment(s) congruent to each segment.
17.
SOLUTION:
Circumcenter D is equidistant from the vertices of
the triangle ABC.
ANSWER:
18.
SOLUTION:
is the perpendicular bisector of . So,
ANSWER:
19.
SOLUTION:
is the perpendicular bisector of . So,
ANSWER:
20.
SOLUTION:
Circumcenter D is equidistant from the vertices of
the triangle ABC.
Therefore,
ANSWER:
Find each measure.
21.AF
SOLUTION:
By the Angle Bisector Theorem, AF = AD = 11.
ANSWER:
11
22.mDBA
SOLUTION:
by the converse of the
Angle Bisector Theorem.
ANSWER:
23.
SOLUTION:
The converse of the Angle Bisector Theorem says
That is,
Solve the equation for x.
Now,
ANSWER:
24.XA
SOLUTION:
By the Angle Bisector Theorem, XA = ZA= 4.
ANSWER:
4
25.mPQS
SOLUTION:
In triangle QRS,
Substitute the known values.
By the converse of the Angle Bisector Theorem,
Therefore
ANSWER:
26.PN
SOLUTION:
Here , by the Angle Bisector Theorem.
Substitute x = 8 in the expression for PN.
ANSWER:
30
CCSS SENSE-MAKINGPoint P is the incenter
of . Find each measure below.
27.PB
SOLUTION:
Use the Pythagorean Theorem in the right triangle
ABP.
ANSWER:
7.1
28.DE
SOLUTION:
is the angle bisector of BytheAngle
Bisector Theorem, PD = PB. Or PD = 7.1.
In triangle PDE,
ANSWER:
13.1
29.m DAC
SOLUTION:
By the Converse of the Angle Bisector Theorem,
istheanglebisectorof . Therefore,
m
ANSWER:
30. DEP
SOLUTION:
By SAS postulate,
So,
ANSWER:
31.INTERIOR DESIGN You want to place a
centerpiece on a corner table so that it is located the
same distance from each edge of the table. Make a
sketch to show where you should place the
centerpiece. Explain your reasoning.
SOLUTION:
Find the point of concurrency of the angle bisectors
of the triangle, the incenter. This point is equidistant
from each side of the triangle.
ANSWER:
Find the point of concurrency of the angle bisectors
of the triangle, the incenter. This point is equidistant
from each side of the triangle.
Determine whether there is enough information
given in each diagram to find the value of x.
Explain your reasoning.
32.
SOLUTION:
There is not enough information to find the value of
x. In order to determine the value of x, we would
need to see markings indicating whether the
segments drawn from the bisecting ray are
perpendicular to each side of the angle being
bisected. Since there are no markings indicating
perpendicularity,noconclusioncanbemade.
ANSWER:
No; we need to know if the segments are
perpendicular to the rays.
33.
SOLUTION:
There is not enough information to find the value of
x. In order to determine the value of x, we would
need to see markings indicating whether the
segments drawn from the ray are congruent to each
other. Since there are no markings indicating
congruence,noconclusioncanbemade.
ANSWER:
No; we need to know whether the perpendicular
segments are congruent to each other.
34.
SOLUTION:
There is not enough information to find the value of
x. In order to determine the value of x, we would
need to see markings indicating whether the segment
bisector is perpendicular to the side of the triangle
that is bisected. Since there are no markings
indicating perpendicularity, no conclusion can be
made.
ANSWER:
No; we need to know if the segment bisector is a
perpendicular bisector.
35.
SOLUTION:
There is not enough information to find the value of
x. According to the Converse of the Perpendicular
Bisector Theorem, we would need to know if the
point of the triangle is equidistant to the endpoints of
the segment. Since we don't know if the hypotenuses
of the two smaller triangles are congruent, we can't
assume that the other side of the big triangle is
bisected.
ANSWER:
No; we need to know whether the hypotenuses of
the triangles are congruent.
36.SOCCER A soccer player P is approaching the
opposing team’s goal as shown in the diagram. To
make the goal, the player must kick the ball between
the goal posts at L and R. The goalkeeper faces the
kicker. He then tries to stand so that if he needs to
dive to stop a shot, he is as far from the left-hand
side of the shot angle as the right-hand side.
a. Describe where the goalkeeper should stand.
Explain your reasoning.
b. Copy . Use a compass and a straightedge
to locate point G, the desired place for the
goalkeeper to stand.
c. If the ball is kicked so it follows the path from P to
R, construct the shortest path the goalkeeper should
take to block the shot. Explain your reasoning.
SOLUTION:
a. The goalkeeper should stand along the angle
bisector of the opponent’s shot angle, since the
distance to either side of the angle is the same along
this line.
b. To determine a point, construct the angle bisector
of ∠P and then mark point Ganywhereon .
c. We can find the shortest distance from point G to
either side of ∠RPL by constructing a line that is
perpendicular to each side of the angle that passes
through point G. The shortest distance to a line from
a point not on the line is the length of the segment
perpendicular to the line from the point
.
ANSWER:
a. The goalkeeper should stand along the angle
bisector of the opponent’s shot angle, since the
distance to either side of the angle is the same along
this line.
b.
c.
The shortest distance to a line from a point not on the
line is the length of the segment perpendicular to the
line from the point.
PROOF Write a two-column proof.
37.Theorem 5.2
Given:
Prove: C and D are on the perpendicular bisector
of
SOLUTION:
As in all proofs, you need to think backwards. What
would you need to do to prove that a segment in a
perpendicular bisector? You can prove it in two parts
- first, that and are perpendicular to each
other and then, that isbisected.Thiswillinvolve
proving two triangles are congruent so that you can
get congruent corresponding parts (CPCTC). Start
by considering which triangles you can make
congruent to each other using the given information.
There are three pairs of triangles in the diagram, so
mark the given information to lead you to the correct
pair. Once you have determined that
, then think about what congruent
corresponding parts you need to make
. To prove that two segments are
perpendicular, consider what CPCTC you need to
choose that would eventually prove that two
segments form a right angle. In addition, which
CPCTC would you choose to prove that a segment is
bisected?
Proof:
Statement(Reasons)
1. (Given)
2. (Congruence of segments is reflexive.)
3. (SSS)
4. (CPCTC)
5. (Congruence of segments is reflexive.)
6. (SAS)
7. (CPCTC)
8. E is the midpoint of . (Definition of midpoint)
9. (CPCTC)
10. formalinearpair.(Definition
of linear pair)
11. aresupplementary.
(Supplementary Theorem)
12. (Definition of
supplementary)
13. (Substitution
Property)
14. (Substitution Property)
15. (Division Property)
16. arerightangles.(Definitionof
right angle)
17. (Definition of Perpendicular lines)
18. istheperpendicularbisectorof .
(Definition of perpendicular bisector)
19. C and D are on the perpendicular bisector of
. (Definition of point on a line)
ANSWER:
Statement(Reasons)
1. (Given)
2. (Congruence of segments is reflexive.)
3. (SSS)
4. (CPCTC)
5. (Congruence of segments is reflexive.)
6. (SAS)
7. (CPCTC)
8. E is the midpoint of . (Definition of midpoint)
9. (CPCTC)
10. formalinearpair.(Definition
of linear pair)
11. aresupplementary.
(Supplementary Theorem)
12. (Definition of
supplementary)
13. (Substitution
Property)
14. (Substitution Property)
15. (DivisionProperty)
16. arerightangles.(Definitionof
right angle)
17. (DefinitionofPerpendicularlines)
18. istheperpendicularbisectorof .
(Definition of perpendicular bisector)
19. C and D are on the perpendicular bisector of
. (Definition of point on a line)
38.Theorem 5.6
Given: , angle bisectors and
Prove:
SOLUTION:
Consider what it means if are
allanglebisectorsandmeetatpointK.Thekeyto
thisproofistheAngleBisectorTheorem.
Proof:
Statement(Reasons)
1. anglebisectors and
(Given)
2. KP = KQ, KQ = KR, KP = KR (Any point on the
angle bisector is equidistant from the sides of the
angle.)
3. KP = KQ = KR(Transitive Property)
ANSWER:
Proof:
Statement(Reasons)
1. anglebisectors and
(Given)
2. KP = KQ, KQ = KR, KP = KR (Any point on the
angle bisector is equidistant from the sides of the
angle.)
3. KP = KQ = KR(Transitive Property)
CCSSARGUMENTSWriteaparagraphproof
of each theorem.
39.Theorem 5.1
SOLUTION:
If we know that is the perpendicular bisector to
, then what two congruent triangle pairs can we
get from this given information? Think about what
triangles we can make congruent to prove that
?
Given: isthe bisectorof
E is a point on CD.
Prove: EA = EB
Proof: isthe ⊥ bisector of Bydefinitionof
bisector, D is the midpoint of Thus,
bytheMidpointTheorem. CDA and
CDB are right angles by the definition of
perpendicular. Since all right angles are congruent,
CDA CDB. Since E is a point on
EDA and EDB are right angles and are
congruent. By the Reflexive Property,
Thus EDA EDB by SAS.
becauseCPCTC,andbydefinitionof
congruence, EA = EB.
ANSWER:
Given: isthe bisectorof
E is a point on CD.
Prove: EA = EB
Proof: isthe ⊥ bisector of Bydefinitionof
bisector, D is the midpoint of Thus,
bytheMidpointTheorem. CDA and
CDB are right angles by the definition of
perpendicular. Since all right angles are congruent,
CDA CDB. Since E is a point on
EDA and EDB are right angles and are
congruent. By the Reflexive Property,
Thus EDA EDB by SAS.
becauseCPCTC,andbydefinitionof
congruence, EA = EB.
40.Theorem 5.5
SOLUTION:
If PD and PE are equidistant to the sides of ,
then they are also perpendicular to those sides.If you
can prove , then consider what
CPCTC would need to be chosen to prove that
is an angle bisector of
Given: BAC
P is in the interior of BAC
PD = PE
Prove: istheanglebisectorof BAC
Proof: Point P is on the interior of BAC of
andPD = PE. By definition of congruence,
and sincethe
distance from a point to a line is measured along the
perpendicular segment from the point to the line.
ADP and AEP are right angles by the definition
of perpendicular lines and and are
right triangles by the definition of right triangles. By
the Reflexive Property, Thus,
byHL. DAP EAP because
CPCTC, and istheanglebisectorof BAC by
the definition of angle bisector.
ANSWER:
Given: BAC
P is in the interior of BAC
PD = PE
Prove: istheanglebisectorof BAC
Proof: Point P is on the interior of BAC of
andPD = PE. By definition of congruence,
. and sincethe
distance from a point to a line is measured along the
perpendicular segment from the point to the line.
ADP and AEP are right angles by the definition
of perpendicular lines and and are
right triangles by the definition of right triangles. By
the Reflexive Property, Thus,
byHL. DAP EAP because
CPCTC, and istheanglebisectorof BAC by
the definition of angle bisector.
COORDINATE GEOMETRY Write an
equation in slope-intercept form for the
perpendicular bisector of the segment with the
given endpoints. Justify your answer.
41.A(–3, 1) and B(4, 3)
SOLUTION:
The slope of the segment AB is or So,the
slope of the perpendicular bisector is
The perpendicular bisector passes through the
midpoint of the segment AB. The midpoint of AB is
or
The slope-intercept form for the equation of the
perpendicular bisector of AB is:
ANSWER:
; The perpendicular bisector bisects
the segment at the midpoint of the segment. The
midpoint is . The slope of the given segment
is , so the slope of the perpendicular bisector is
.
42.C(–4, 5) and D(2, –2)
SOLUTION:
The slope of the segment CD is or So,
the slope of the perpendicular bisector is
The perpendicular bisector passes through the
midpoint of the segment CD. The midpoint of CD is
or
The slope-intercept form for the equation of the
perpendicular bisector of CD is:
ANSWER:
; The perpendicular bisector bisects the
segment at the midpoint of the segment. The
midpoint is .Theslopeofthegivensegment
is ,sotheslopeoftheperpendicularbisector
is .
43.PROOF Write a two-column proof of Theorem 5.4.
SOLUTION:
The key to this proof is to think about how to get
. Use your given information to
get congruent corresponding parts. Remember that
an angle bisector makes two anglescongruent.
Given: bisects QPR. and
Prove:
Proof:
Statements (Reasons)
1. bisects QPR,and .
(Given)
2. (Definition of angle bisector)
3. PYX and PZX are right angles. (Definition of
perpendicular)
4. (Right angles are congruent.)
5. (Reflexive Property)
6.
7. (CPCTC)
ANSWER:
Given: bisects QPR. and
.
Prove:
Proof:
Statements (Reasons)
1. bisects QPR,and .
(Given)
2. (Definition of angle bisector)
3. PYX and PZX are right angles. (Definition of
perpendicular)
4. (Right angles are congruent.)
5. (Reflexive Property)
6.
7. (CPCTC)
44.GRAPHIC DESIGN Mykia is designing a pennant
for her school. She wants to put a picture of the
school mascot inside a circle on the pennant. Copy
the outline of the pennant and locate the point where
the center of the circle should be to create the largest
circle possible. Justify your drawing.
SOLUTION:
We need to find the incenter of the triangle by finding
the intersection point of the angle bisectors. To make
the circle, place your compass on the incenter, mark
a radius that is perpendicular to each side (the
shortest length) and make a circle.When the circle is
as large as possible, it will touch all three sides of the
pennant.
ANSWER:
When the circle is as large as possible, it will touch
all three sides of the pennant. We need to find the
incenter of the triangle by finding the intersection
point of the angle bisectors.
COORDINATE GEOMETRY Find the
coordinates of the circumcenter of the triangle
with the given vertices. Explain.
45.A(0, 0), B(0, 6), C(10, 0)
SOLUTION:
Graph the points.
Since the circumcenter is formed by the
perpendicular bisectors of each side of the triangle,
we need to draw the perpendicular bisectors of the
two legs of the triangle and see where they intersect.
The equation of a line of one of the perpendicular
bisectors is y = 3 because it is a horizontal line
throughpoint(0,3),themidpointoftheverticalleg.
The equation of a line of another perpendicular
bisector is x = 5, because it is a vertical line through
thepoint(5,0),themidpointofthehorizontalleg.
These lines intersect at (5, 3), because it will have
the x-value of the line x=5 and the y-value of the line
y=3.
The circumcenter is located at (5, 3).
ANSWER:
The equation of a line of one of the perpendicular
bisectors is y = 3. The equation of a line of another
perpendicular bisector is x = 5. These lines intersect
at (5, 3). The circumcenter is located at (5, 3).
46.J(5, 0), K(5, –8), L(0, 0)
SOLUTION:
Since the circumcenter is formed by the
perpendicular bisectors of each side of the triangle,
we need to draw the perpendicular bisectors of the
two legs of the triangle and see where they intersect.
The equation of a line of one of the perpendicular
bisectors is y = –4 because it is a horizontal line
through point (5, –4),themidpointoftheverticalleg.
The equation of a line of another perpendicular
bisector is x = 2.5, because it is a vertical line through
thepoint(2.5,0),themidpointofthehorizontalleg.
These lines intersect at (2.5, –4), because it will have
the x-value of the line x = 2.5 and the y-value of the
line y = –4.
The circumcenter is located at (2.5, –4).
ANSWER:
The equation of a line of one perpendicular bisector
is y = –4. The equation of a line of another
perpendicular bisector is x = 2.5. These lines
intersect at (2.5, –4). The circumcenter is located at
(2.5, –4).
47.LOCUS Consider .Describethesetofall
points in space that are equidistant from C and D.
SOLUTION:
Since we are considering all points in space, we
need to consider more than just the perpendicular
bisector of CD. The solution is a plane that is
perpendicular to the plane in which liesand
bisects .
ANSWER:
a plane perpendicular to the plane in which lies
and bisecting .
48.ERROR ANALYSIS Claudio says that from the
information supplied in the diagram, he can conclude
that K is on the perpendicular bisector of .
Caitlyn disagrees. Is either of them correct? Explain
your reasoning.
SOLUTION:
Caitlyn is correct; Based on the markings, we only
know that J is the midpoint of . We don't know if
is perpendicular to .
ANSWER:
Caitlyn; K is only on the perpendicular bisector of
if , but we are not given this
information in the diagram.
49.OPEN ENDED Draw a triangle with an incenter
located inside the triangle but a circumcenter located
outside. Justify your drawing by using a straightedge
and a compass to find both points of concurrency.
SOLUTION:
Knowing that the incenter of a triangle is always
found inside any triangle, the key to this problem is to
think about for which type of triangle the
circumcenter would fall on the outside. Experiment
with acute, right and obtuse triangles to find the one
thatwillwork.
ANSWER:
Sample answer:
CCSS ARGUMENTS Determine whether each
statement is sometimes, always, or never true.
Justify your reasoning using a counterexample
or proof.
50.The angle bisectors of a triangle intersect at a point
that is equidistant from the vertices of the triangle.
SOLUTION:
It is the circumcenter, formed by the perpendicular
bisectors, that is equidistant to the vertices of a
triangle. Consider for what type of triangle the
incenter, formed by the angle bisectors, would be the
same location as the circumcenter. Sometimes; if the
triangle is equilateral, then this is true, but if the
triangle is isosceles or scalene, the statement is false.
.
JQ = KQ = LQ
ANSWER:
Sometimes; if the triangle is equilateral, then this is
true, but if the triangle is isosceles or scalene, the
statement is false.
.
JQ = KQ = LQ
51.In an isosceles triangle, the perpendicular bisector of
the base is also the angle bisector of the opposite
vertex.
SOLUTION:
Always. It is best to show this through a proof that
would show that, given an isosceles triangle divided
by a perpendicular bisector, that it creates two
congruent triangles. Therefore, by CPCTC, the two
angles formed by the perpendicular bisector are
congruentandformananglebisector.
Given: is isosceles with legs and ;
is the bisector of .
Prove: is the angle bisector of ABC.
Proof:
Statements (Reasons)
1. isisosceleswithlegs and . (Given)
2. (Definition of isosceles triangle)
3. is the bisector of .(Given)
4. D is the midpoint of . (Definition of segment
bisector)
5. (Definition of midpoint)
6. (Reflexive Property)
7.
8.
9. is the angle bisector of ABC. (Definition of
angle bisector)
ANSWER:
Always.
Given: is isosceles with legs and ;
is the bisector of .
Prove: is the angle bisector of ABC.
Proof:
Statements (Reasons)
1. isisosceleswithlegs and . (Given)
2. (Def. of isosceles )
3. is the bisector of .(Given)
4. D is the midpoint of . (Def. of segment
bisector)
5. (Def. of midpoint)
6. (Reflexive Property)
7.
8.
9. is the angle bisector of ABC . (Def.
bisector)
CHALLENGE Write a two-column proof for
each of the following.
52.Given: Plane Yisaperpendicular
bisector of .
Prove:
SOLUTION:
A plane that is a perpendicular bisector of a segment
behaves the same way as a segment that is a
perpendicular bisector. It creates two right angles
and two congruent corresponding segments. Use
them to prove two triangles congruent in this
diagram.
Proof:
Statements (Reasons)
1. Plane Y is a perpendicular bisector of . (Given)
2. DBA and CBA are right angles,
(Definitionof bisector)
3. (Right angles are congruent.)
4. (Reflexive Property)
5. (SAS)
6. (CPCTC)
ANSWER:
Proof:
Statements (Reasons)
1. Plane Y is a perpendicular bisector of . (Given)
2. DBA and CBA are rt. s, (Def.
of bisector)
3. (Right angles are congruent.)
4. (Reflexive Property)
5. (SAS)
6. (CPCTC)
53.Given: Plane Z is an angle bisector of
.
Prove:
SOLUTION:
A plane that is an angle bisector of a segment
behaves the same way as a segment that is an angle
bisector. It creates two congruent angles. Use these
angles, along with the other given and the diagram, to
prove two triangles congruent.
Proof:
Statements (Reasons)
1. Plane Z is an angle bisector of KJH;
(Given)
2. (Definition of angle bisector)
3. (Reflexive Property)
4.
5. (CPCTC)
ANSWER:
Proof:
Statements (Reasons)
1. Plane Z is an angle bisector of KJH;
(Given)
2. (Definitionofanglebisector)
3. (ReflexiveProperty)
4.
5. (CPCTC)
54.WRITING IN MATH Compare and contrast the
perpendicular bisectors and angle bisectors of a
triangle. How are they alike? How are they
different? Be sure to compare their points of
concurrency.
SOLUTION:
Start by making a list of what the different bisectors
do. Consider what they bisect, as well as the point of
concurrency they form. Discuss the behaviors of
these points of concurrency in different types of
triangles. Are they ever found outside the triangle?
Insidethetriangle?Onthetriangle?
The bisectors each bisect something, but the
perpendicular bisectors bisect segments while angle
bisectors bisect angles. They each will intersect at a
point of concurrency. The point of concurrency for
perpendicular bisectors is the circumcenter. The
point of concurrency for angle bisectors is the
incenter. The incenter always lies in the triangle,
while the circumcenter can be inside, outside, or on
the triangle.
ANSWER:
The bisectors each bisect something, but the
perpendicular bisectors bisect segments while angle
bisectors bisect angles. They each will intersect at a
point of concurrency. The point of concurrency for
perpendicular bisectors is the circumcenter. The
point of concurrency for angle bisectors is the
incenter. The incenter always lies in the triangle,
while the circumcenter can be inside, outside, or on
the triangle.
55.ALGEBRA An object is projected straight upward
with initial velocity v meters per second from an
initial height of s meters. The distance d in meters
the object travels after t seconds is given by d = –
10t2+ vt + s. Sherise is standing at the edge of a
balcony 54 meters above the ground and throws a
ball straight up with an initial velocity of 12 meters
per second. After how many seconds will it hit the
ground?
A 3 seconds
B 4 seconds
C 6 seconds
D 9 seconds
SOLUTION:
Substitute the known values in the distance
expression.
Solve the equation for t.
Discard the negative value, as it has no real world
meaning in this case.
The correct choice is A.
ANSWER:
A
56.SAT/ACT For
F x + 12
G x + 9
H x + 3
J x
K 3
SOLUTION:
The correct choice is K.
ANSWER:
K
57.A line drawn through which of the following points
would be a perpendicular bisector of ?
A T and K
B L and Q
C J and R
D S and K
SOLUTION:
The line passes through the points K and S will be a
perpendicular bisector of
The correct choice is D.
ANSWER:
D
58.SAT/ACT For x≠–3,
Fx + 12
Gx + 9
Hx + 3
J x
K3
SOLUTION:
Simplify the expression by first removing the GCF.
Therefore, the correct choice is K.
ANSWER:
K
Name the missing coordinate(s) of each
triangle.
59.
SOLUTION:
The line drawn through the point L is the
perpendicular bisector of Therefore,x-
coordinate of L is x-coordinate of the midpoint of JK.
The x-coordinate of L is ora.
Thus, the coordinates of L is (a, b).
ANSWER:
L(a, b)
60.
SOLUTION:
Since AC = AB, the distance from the origin to point
B and C is the same. They are both a length away
from the origin. The coordinate of C is (a, 0) because
it is on the x-axis and is a units away from the
origin.
ANSWER:
C(a, 0)
61.
SOLUTION:
Since the triangle is an isosceles triangle, S and T are
equidistant from O. So, the coordinates of S are (–
2b, 0). R lies on the y-axis. So, its x-coordinate is 0.
The coordinates of R are (0, c)
ANSWER:
S(–2b, 0) and R(0, c)
COORDINATE GEOMETRY Graph each pair
of triangles with the given vertices. Then
identify the transformation and verify that it is a
congruence transformation.
62.A(–2, 4), B(–2, –2), C(4, 1);
R(12, 4), S(12, –2), T(6, 1)
SOLUTION:
is a reflection of .
Findthelengthofeachside.
by SSS.
ANSWER:
is a reflection of ; AB = 6, BC = ,
AC = , TR = , RS = 6, TS= .
by SSS.
63.J(–3, 3), K(–3, 1), L(1, 1);
X(–3, –1), Y(–3, –3), Z(1, –3)
SOLUTION:
is a translation of .
Find the length of each side.
by SSS.
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5-1 Bisectors of Triangles
