Transcript B - Ms. Williams – Math

Congruent Figures
GEOMETRY LESSON 4-1
(For help, go to page 24.)
Solve each equation.
1. x + 6 = 25
2. x + 7 + 13 = 33
3. 5x = 540
4. x + 10 = 2x
5. For the triangle at the right, use the Triangle
Angle-Sum Theorem to find the value of y.
4-1
Congruent Figures
GEOMETRY LESSON 4-1
Solutions
1. Subtract 6 from both sides: x = 19
2. Combine like terms: x + 20 = 33; subtract 20 from both sides: x = 13
3. Divide both sides by 5: x = 108
4. Subtract x from both sides of x + 10 = 2x: 10 = x, or x = 10
5. y + 40 + 90 = 180; combine like terms: y + 130 = 180;
subtract 130 from both sides: y = 50
4-1
Congruent Figures
GEOMETRY LESSON 4-1
ABC
QTJ. List the congruent corresponding
parts.
List the corresponding vertices in the same order.
Angles: A
Q
B
T
C
J
List the corresponding sides in the same order.
Sides: AB
QT
BC
TJ
AC
4-1
QJ
Congruent Figures
GEOMETRY LESSON 4-1
XYZ
KLM, mY = 67, and mM = 48. Find mX.
Use the Triangle Angle-Sum Theorem and the definition of congruent
polygons to find mX.
mX + mY + mZ = 180
Triangle Angle-Sum Theorem
mZ = mM
Corresponding angles of congruent
triangles that are congruent
mZ = 48
Substitute 48 for mM.
mX + 67 + 48 = 180
mX + 115 = 180
mX = 65
Substitute.
Simplify.
Subtract 115 from each side.
4-1
Congruent Figures
GEOMETRY LESSON 4-1
Can you conclude that
ABC
CDE in the figure below?
List corresponding vertices in the same order.
If
ABC
CDE, then BAC
DCE.
The diagram above shows BAC
DEC, not DCE.
Corresponding angles are not necessarily congruent, therefore you
cannot conclude that ABC
CDE.
4-1
Congruent Figures
GEOMETRY LESSON 4-1
Show how you can conclude that CNG
DNG. List
statements and reasons.
Congruent triangles have three congruent corresponding
sides and three congruent corresponding angles.
Examine the diagram, and list the congruent corresponding
parts for CNG and DNG.
a. CG DG
b. CN DN
c. GN GN
d. C D
e. CNG DNG
f. CGN DGN
g. CNG
DNG
Given
Given
Reflexive Property of Congruence
Given
Right angles are congruent.
If two angles of one triangle are congruent to two angles
of another triangle, then the third angles are congruent.
(Theorem 4-1.)
Definition of triangles
4-1
Congruent Figures
GEOMETRY LESSON 4-1
Pages 182-185 Exercises
1.
2.
CAB
DAB;
C
D;
ABC
ABD;
AC AD; AB AB
CB DB
GEF
JHI;
GFE
JIH;
EGF
HJI;
GE JH; EF HI
FG IJ
5. ML
13. E, K, G, N
6.
B
14. PO SI; OL ID;
LY DE; PY SE
7.
C
15.
P
L
8.
J
9.
KJB
16. 33 in.
10.
CLM
17. 54 in.
3. BK
11.
JBK
18. 105
4. CM
12.
MCL
19. 77
4-1
S;
D;
O
Y
I;
E
Congruent Figures
GEOMETRY LESSON 4-1
20. 36 in.
27. Yes; all corr. sides and
s are
.
21. 34 in.
28. a. Given
b. If || lines, then alt. int.
.
s are
22. 75
23. 103
24. Yes; RTK
UTK, R
U (Given)
RKT
UKT
If two s of a are to two s of another , the third
s are
. TR TU, RK UK (Given)
TK TK (Reflexive Prop. of ) TRK
TUK
s )
(Def. of
25. No; the corr. sides are not
.
26. No; corr. sides are not necessarily
.
4-1
c. Given
d. If 2 s of one are
to two s of another
then 3rd s are .
e. Reflexive Prop. of
f. Given
g. Def. of
s
,
Congruent Figures
GEOMETRY LESSON 4-1
29. A and H; B and G;
C and E; D and F
30. x = 15; t = 2
36. Answers may vary.
Sample: It is important
that PACH OLDE for
the patch to completely
fill the hole.
31. 5
32. m
A=m
D = 20
33. m B = m E = 21
37. Answers may vary.
Sample: She could
arrange them in a neat
pile and pull out the
ones of like sizes.
34. BC = EF = 8
38.
JYB
XCH
39.
BCE
ADE
40.
TPK
TRK
35. AC = DF = 19
4-1
41.
JLM
JLM
NRZ;
ZRN
42. Answers may vary.
Sample: The die is a
mold that is used to
make items that are
all the same size.
43. Answers may vary.
Sample:
TKR
MJL:
TK MJ; TR ML;
KR JL;
TKR
MJL;
TRK
MLJ;
KTR
JML
Congruent Figures
GEOMETRY LESSON 4-1
44. a. Given
b. If || lines, then alt.
int. s are .
c. If || lines, then alt.
int. s are .
d. Vertical
s
are
e. Given
f. Given
46. KL = 4; LM = 3; KM = 5
g. Def. of segment
bisector
h. Def. of
.
45. Answers may vary.
Sample: Since the sum
of the s of a is 180,
and if 2 s of one are
the same as 2 s of a
second , then their
sum subtracted from
180 has to be the
same.
47. 2; either (3, 1) or (3, –7)
s
4-1
48. a. 15
b.
Congruent Figures
GEOMETRY LESSON 4-1
49. 1.5
54.
50. 4.25
51. 40
52. 72
55. 100
53. Answers may vary.
Sample:
56. RS = PQ
57.
1
58. 12
59. AB
GH
4-1
Congruent Figures
GEOMETRY LESSON 4-1
In Exercises 1 and 2, quadrilateral WASH quadrilateral NOTE.
1. List the congruent corresponding parts.
WA NO, AS OT, SH TE, WH NE;
W N, A O, S T, H E
2. mO = mT = 90 and mH = 36. Find mN.
144
3. Write a statement of triangle congruence.
Sample:
DFH
ZPR
4. Write a statement of triangle congruence.
Sample:
ABD
CDB
5. Explain your reasoning in Exercise 4 above.
Sample: Two pairs of corresponding sides and two
pairs of corresponding angles are given. C A
because all right angles are congruent. BD BD by the
Reflexive Property of . ABD
CDB by the
definition of congruent triangles.
4-1
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
(For help, go to Lesson 2-5.)
What can you conclude from each diagram?
1.
2.
3.
4-2
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
Solutions
1. According to the tick marks on the sides, AB
tick marks on the angles, C F.
DE. According to the
2. The two triangles share a side, so PR PR. According to the tick
marks on the angles, QPR SRP and Q S.
3. According to the tick marks on the sides, TO NV. The tick marks on
the angles show that M S. Since MO || VS, by the Alternate
Interior Angles Theorem MON SVT. Since OV OV by the
Reflexive Property, you can use the Segment Addition Property to
show TV NO.
4-2
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
Given: M is the midpoint of XY, AX
Prove: AMX
AMY
AY
Write a paragraph proof.
Copy the diagram. Mark the congruent sides.
You are given that M is the midpoint of XY, and AX AY.
Midpoint M implies MX MY. AM AM by the Reflexive
Property of Congruence, so AMX
AMY by the SSS
Postulate.
4-2
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
AD
BC. What other information do you need to prove
ADC
BCD?
It is given that AD BC. Also, DC CD by the Reflexive Property of
Congruence. You now have two pairs of corresponding congruent sides.
Therefore:
Solution 1: If you know AC
by SSS.
BD, you can prove
Solution 2: If you know ADC
by SAS.
ADC
BCD, you can prove
4-2
BCD
ADC
BCD
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
Given: RSG RSH, SG SH
From the information given, can you prove
RSG
RSH? Explain.
Copy the diagram. Mark what is given on the diagram.
It is given that RSG RSH and SG SH.
RS RS by the Reflexive Property of Congruence.
Two pairs of corresponding sides and their included angles
are congruent, so RSG
RSH by the SAS Postulate.
4-2
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
Pages 189-192 Exercises
3. SAS
6. Yes; AC DB (Given);
10. U,
AE CE and BE DE
(Def. of midpt.); AEB
11. WU
s
CED (vert. are )
AEB
CED by SAS. 12. X
4. SSS
7. a. Given
1. SSS
2. cannot be proved
5. Yes; OB OB by
Refl. Prop.; BOP
BOR since all rt.
are ; OP OR
(Given); the s are
by SAS.
b. Reflexive
c. JKM
d. LMK
s
V
13. XZ, YZ
14. LG
15.
T
or RS
MN
V
WU
8. WV, VU
16. DC
9.
W
4-2
CB
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
17. additional information
not needed
18. Yes; ACB
by SAS.
19. Yes; PVQ
by SSS.
EFD
22.
ANG
LOM
KLJ
MON; SSS
29. No; you would need
H
K or GI JL.
24. Not possible; need
H
P or DY TK.
30. yes; SAS
25.
JEF
JEF
31.
26.
BRT
BRS; SSS
27.
PQR
NMO; SAS
STR
20. No; need YVW
ZVW or YW ZW.
21. Yes; NMO
by SAS.
23.
SVF or
SFV; SSS
28. No; even though the s
are , the sides may
not be.
RWT; SAS
4-2
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
32.
34. Answers may vary.
Sample:
35. (continued)
b.
35. a–b. Answers may
vary. Sample:
33. a.
b.
c.
d.
Vertical s are
Given
Def. of midpt.
Given
e.
Def. of midpt.
f.
SAS
.
a. wallpaper designs;
ironwork on a bridge;
highway warning
36.
signs
produce a wellbalanced, symmetric
appearance. In
s
construction,
enhance designs.
Highway warning signs
are more easily
identified if they are .
s
ISP
ISP
by SAS.
PSO;
OSP
37. IP PO; ISP
OSP by SSS.
4-2
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
38. Yes; ADB
CBD
by SAS; ADB
DBC because if ||
lines, then alt. int. s
are .
41. 1. FG || KL (Given)
2.
GFK
FKL (If ||
lines, then alt. int. s
are .)
3. FG
39. Yes; ABC
CDA
by SAS; DAC
ACB because if ||
lines, then alt. int. s
are .
KL (Given)
4. FK FK (Reflexive
Prop. of )
5.
FGK
(SAS)
40. No; ABCD could be a
square with side 5
and EFGH could be a
polygon with side 5
but no rt. s .
4-2
KLF
42. AE and BD bisect
each other, so
AC CE and
BC CD.
ACB
DCE because svert.
are .
ACB
ECD by
SAS.
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
43.
45. D
51. FG
46. G
52.
47. C
48. [2] a. AB AB;
Reflexive Prop.
of
44. AM MB because M
is the midpt. of AB.
B
AMC
because all right s are
. CM DB is given.
AMC
MBD by
SAS.
b. No; AB is not a
corr. side.
[1] one part correct
49.
E
50. AB
4-2
C
53. The product of the
slopes of two lines is –
1 if and only if the
lines are .
54. If x = 2, then 2x = 4.
If 2x = 4, then x = 2.
55. If 2x = 6, then x = 3.
The statement and the
converse are both
true.
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
56. If x2 = 9, then x = 3.
The statement is true
but the converse is
false.
4-2
Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
1. In VGB, which sides include B? BG and BV
2. In STN, which angle is included between NS and TN? N
3. Which triangles can you prove congruent?
Tell whether you would use the SSS or SAS Postulate.
APB
XPY; SAS
4. What other information do you need to prove
DWO
DWG?
If you know DO DG, the triangles are by SSS;
if you know DWO DWG, they are by SAS.
5. Can you prove SED
BUT from the information given?
Explain.
No; corresponding angles are not between
corresponding sides.
4-2
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
(For help, go to the Lesson 4-2.)
In
JHK, which side is included between the given pair of angles?
1. J and H
2. H and K
In
NLM, which angle is included between the given pair of sides?
3. LN and LM
4. NM and LN
Give a reason to justify each statement.
5. PR PR
6. A
4-3
D
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Solutions
1. JH
2. HK
3. L
4. N
5. By the Reflexive Property of Congruence, a segment is congruent to
itself.
6. By the Triangle Angle-Sum Theorem, the sum of the angles of any
triangle is 180. If mC = mF = x and mB = mE = y, then
mA = 180 – x – y = mD, so A = D.
4-3
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Suppose that F is congruent to C and I is not congruent to C. Name
the triangles that are congruent by the ASA Postulate.
The diagram shows N
If F
C, then F
Therefore,
FNI
A
C
CAT
D and FN
G
GDO by ASA.
4-3
CA
GD.
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Write a paragraph proof.
Given: A
Prove:
B, AP
APX
It is given that A
APX
BP
BPY
B and AP
BP.
BPY by the Vertical Angles Theorem.
Because two pairs of corresponding angles and
their included sides are congruent, APX
BPY
by ASA.
4-3
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Write a Plan for Proof that uses AAS.
Given: B
Prove:
D, AB || CD
ABC
CDA
Because AB || CD, BAC
Interior Angles Theorem.
DCA by the Alternate
Then ABC
CDA if a pair of corresponding
sides are congruent.
By the Reflexive Property, AC
ABC
CDA by AAS.
4-3
AC so
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Write a two-column proof that uses AAS.
Given: B D, AB || CD
Prove: ABC
CDA
Statements
Reasons
1. B
1. Given
D, AB || CD
2. BAC
3. AC
4.
DCA
CA
ABC
2. If lines are ||, then alternate interior angles are .
3. Reflexive Property of Congruence
CDA
4. AAS Theorem
4-3
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Pages 197-201 Exercises
1.
PQR
VXW
2.
ACB
EFD
9. AAS
3. RS
4.
N and
8. a. Reflexive
b. ASA
O
5. yes
6. not possible
13. a. UWV
b. UW
c. right
d. Reflexive
10. ASA
14.
B
D
11. not possible
12.
FDE
DFE
GHI;
HGI
7. yes
15. MU
UN
16. PQ
QS
17.
4-3
WZV
WZY
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
18. a. Vert. s are
b. Given
.
c. TQ QR
d. AAS
19.
20.
PMO
UTS
21.
ZVY
22.
TUX
24.
TXU
ODE; ASA
25. The s are not
s are
because the
not included s .
NMO; ASA 26. Yes; if 2 s of a are
RST; AAS
to 2 s of another ,
then the 3rd s are .
WVY; AAS
So, an AAS proof can
be rewritten as an ASA
DEO; AAS
proof.
23. The are not
s
because
no sides are
.
4-3
27. a. SRP
b. PR
c. alt. int.
d. PR
e. Reflexive
28. a. Given
b. Def. of bis.
c. Given
d. Reflexive Prop.
of
e. AAS
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
29. a. Def. of
b. All right
s
are
.
c. QTP
STR
d. Def. of midpt.
e. AAS
30.
31. Yes; by AAS since
MON
QOP.
34. Yes; by ASA since
BDH
FDH by
def. of bis. and
DH DH by the
Reflexive Prop. of .
32. Yes; by AAS since
FGJ
HJG
because when lines are
35. Answers may vary.
||, then alt. int. s are
Sample:
and GJ GJ by the
Reflexive Prop. of .
33. Yes; by ASA, since
EAB
DBC
because || lines have
corr. s .
4-3
36. a. Check students’
work.
b. Answers may vary;
most likely ASA.
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
37.
38.
AEB
BEC
ABC
BAD
AEB
BEC
ABC
ABD
BAD
ABD
CBA
BCD
39. They are
ASA.
CED,
DEA,
CDA,
DCB
CED,
DEA,
CDA,
DCA,
DCB,
DCB,
DAB,
ADC
44. [2] a.
RPQ
SPQ,
RQP
SQP
(Def. of bisector)
40. 13
20
41.
b. ASA
[1] one part correct
42. D
43. F
bisectors;
4-3
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
45. [4] a. Def. of midpt.
b. Yes; JLM
KGM because they are alt.
int. s of || lines, and LMJ
GMK because
vertical s are . So the s are by ASA.
c. Yes; if two s of one are to 2 s of another
, the third s are .
[3] incorrect
s
for part b or c, but otherwise correct
[2] correct conclusions but incomplete explanations
for parts b and c
[1] at least one part correct
46.
ONL
MLN; SAS
47. not possible
4-3
48. AC and CB
49. If corr. s are , then
the lines are ||.
50. 56 photos
51. 36 photos
52. 60% more paper
Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
1. Which side is included between R and F in FTR?
2. Which angles in STU include US? S and U
RF
Tell whether you can prove the triangles congruent by ASA or AAS. If you
can, state a triangle congruence and the postulate or theorem you used. If
not, write not possible.
3.
4.
GHI
AAS
PQR
5.
not possible
4-3
ABX
AAS
ACX
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
(For help, go to the Lesson 4-1.)
In the diagram, JRC
HVG.
1. List the congruent
corresponding angles.
2. List the congruent
corresponding sides.
You are given that TIC
LOK.
3. List the congruent
corresponding angles.
4. List the congruent
corresponding sides.
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
Solutions
1. In the triangle congruence statement, the corresponding vertices are
listed in the same order. So, J H, R V, and C G.
2. In the triangle congruence statement, the corresponding vertices are
listed in the same order. So, JR HV, RC VG, and JC HG.
3. In the triangle congruence statement, the corresponding vertices are
listed in the same order. So, T L, I O, and C K.
4. In the triangle congruence statement, the corresponding vertices are
listed in the same order. So, TI LO, IC OK, and TC LK.
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
What other congruence statements can you prove from the
diagram, in which SL SR, and 1 2 are given? SC SC by the
Reflexive Property of Congruence, and LSC
RSC by SAS.
3 4 by corresponding parts of congruent triangles are congruent.
When two triangles are congruent,
you can form congruence statements
about three pairs of corresponding angles
and three pairs of corresponding sides.
List the congruence statements.
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
(continued)
Sides:
Angles:
SL
SC
CL
1
3
CLS
SR
SC
CR
Given
Reflexive Property of Congruence
Other congruence statement
2
4
Given
Corresponding Parts of
Congruent Triangles
CRS
Other congruence statement
In the proof, three congruence statements are used, and one congruence
statement is proven. That leaves two congruence statements remaining that
also can be proved:
CLS CRS
CL CR
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
The Given states that DEG and DEF are right angles.
What conditions must hold for that to be true?
DEG and DEF are the angles the officer makes with the ground.
So the officer must stand perpendicular to the ground, and the ground
must be level.
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
Pages 204-208 Exercises
1.
PSQ
SPR;
SQ RP; PQ SR
5. They are ; the s are
by AAS, so all corr. ext.
s are also
.
9.
SPT
OPT by
SAS because TP TP
by Reflexive Prop. of
; S
O by
CPCTC.
10.
PNK
MNL by
SAS because KNP
LNM by vert. s are
; KP LM by
CPCTC.
2. AAS; ABC
EBD;
A
E; CB DB; 6. a. SSS
DE CA by CPCTC
b. CPCTC
3. SAS; KLJ
OMN; 7.
ABD
CBD by ASA
K
O; J
N;
because BD BD by
KJ ON by CPCTC
Reflexive Prop. of ;
AB CB by CPCTC.
4. SSS; HUG
BUG;
H
B; HUG
8.
MOE
REO by
11. CYT
BUG; UGH
SSS because OE OE
AAS; CT
UGB by CPCTC
by Reflexive Prop. of ;
CPCTC.
M
R by CPCTC.
4-4
RYP by
RP by
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
12.
ATM
RMT by
SAS because ATM
RMT by alt. int. s
are ; AMT
RTM by CPCTC.
13. Yes; ABD
CBD
by SSS so A
C
by CPCTC.
14. a. Given
b. Given
c. Reflexive Prop.
of
d. AAS
15.
PKL
QKL by def. 18. The s are by SAS
of bisect, and KL KL
so the distance
by Reflexive Prop. of ,
across the sinkhole is
so the s are by SAS.
26.5 yd by CPCTC.
16. KL KL by Reflexive
Prop. of ; PL LQ by
Def. of bis.; KLP
KLQ by Def. of ; the
s are
by SAS.
17.
KLP
KLQ
because all rt s are ;
KL KL by Reflexive
Prop. of ; and PKL
QKL by def. of bisect;
the s are by ASA.
4-4
19. a. Given
b. Def. of
c. All right
s
are
.
d. Given
e. Def. of segment
bis.
f. Reflexive Prop.
of
g. SAS
h. CPCTC
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
20.
ABX
ACX by
SSS, so BAX
CAX by CPCTC.
Thus AX bisects
BAC by the def. of
bisector.
21. Prove ABE
CDF
by SAS since AE
FC by subtr.
22. Prove KJM
QPM
by ASA since P
J and K
Q by
alt. int. s are .
24. BA BC is given; BD BD by the Reflexive Prop.
of and since BD bisects ABC, ABD
CBD
by def. of an bisector; thus, ABD
CBD by
SAS; AD DC by CPCTC so BD bisects AC by def.
of a bis.; ADB
CDB by CPCTC and ADB
and CDB are suppl.; thus, ADB and CDB are
right s and BD AC by def. of .
25. a. AP
PB; AC
BC
b. The diagram is constructed in such a way that
the s are by SSS. CPA
CPB by
CPCTC. Since these s are and suppl., they are
right s . Thus, CP is to .
23. e or b, e or b, d, c, f, a
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
26. 1. PR || MG; MP || GR (Given)
30. D
2. Draw PG. (2 pts. determine a line.)
31. B
3.
32. C
4.
RPG
PGM and
then alt. int. s are .)
RGP
GPM (If || lines,
PGM
GPR (ASA) A similar proof can be
written if diagonal RM is drawn.
33. [2] a. KBV
yes; SAS
KBT;
b. CPCTC
27. Since PGM
GPR (or PMR
GRM), then
PR MG and MP GR by CPCTC.
[1] one part correct
28. C
34. ASA
29. C
35. AAS
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
36. 95; 85
37. The slope of line m is
the same as the slope
of line n.
38. not possible
39. not possible
4-4
Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
1. What does “CPCTC” stand for?
Corresponding parts of congruent triangles are congruent.
Use the diagram for Exercises 2 and 3.
2. Tell how you would show ABM
ACM.
You are given two pairs of s and AM AM by
the Reflexive Prop., so ABM
ACM by ASA.
3. Tell what other parts are congruent by CPCTC.
AB AC, BM CM, B C
Use the diagram for Exercises 4 and 5.
4. Tell how you would show RUQ
TUS.
You are given a pair of s and a pair of sides and
RUQ TUS because vertical angles are , so
RUQ
TUS by AAS.
5. Tell what other parts are congruent by CPCTC.
RQ
TS, UQ
US, R
T
4-4
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
(For help, go to the Lesson 3-3.)
1. Name the angle opposite AB.
2. Name the angle opposite BC.
3. Name the side opposite A.
4. Name the side opposite C.
5. Find the value of x.
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Solutions
1. The angle opposite AB is the angle whose side is not AB: C
2. The angle opposite BC is the angle whose side is not BC: A
3. The side opposite A is the side that is not part of A: BC
4. The side opposite C is the side that is not part of C: BA
5. By the Triangle Exterior Angle Theorem, x = 75 + 30 = 105°.
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Examine the diagram below. Suppose that you draw XB YZ.
Can you use SAS to prove XYB
XZB? Explain.
It is given that XY XZ.
By the definition of perpendicular, XBY = XBZ.
By the Reflexive Property of Congruence, XB XB.
However, because the congruent angles are not included between the
congruent corresponding sides, the SAS Postulate does not apply.
You cannot prove the triangles congruent using SAS.
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Explain why
ABC is isosceles.
ABC and XAB are alternate interior angles
formed by XA, BC, and the transversal AB. Because
XA || BC, ABC XAB.
The diagram shows that XAB ACB. By the
Transitive Property of Congruence, ABC ACB.
You can use the Converse of the Isosceles Triangle
Theorem to conclude that AB AC.
By the definition of an isosceles triangle,
isosceles.
4-5
ABC is
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Suppose that mL = y. Find the values of x and y.
MO
LN
The bisector of the vertex angle of an
isosceles triangle is the perpendicular
bisector of the base.
x = 90
Definition of perpendicular
mN = mL Isosceles Triangle Theorem
mL = y
Given
mN = y
Transitive Property of Equality
mN + mNMO + mMON = 180 Triangle Angle-Sum Theorem
y + y + 90 = 180 Substitute.
2y + 90 = 180 Simplify.
2y = 90
Subtract 90 from each side.
y = 45
Divide each side by 2.
Therefore, x = 90 and y = 45.
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Suppose the raised garden bed is a regular hexagon.
Suppose that a segment is drawn between the endpoints of the angle
marked x. Find the angle measures of the triangle that is formed.
Because the garden is a regular hexagon, the sides
have equal length, so the triangle is isosceles.
By the Isosceles Triangle Theorem, the unknown
angles are congruent.
Example 4 found that the measure of the angle
marked x is 120. The sum of the angle measures of
a triangle is 180.
If you label each unknown angle y, 120 + y + y = 180.
120 + 2y = 180
2y = 60
y = 30
So the angle measures in the triangle are 120, 30 and 30.
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Pages 213-216 Exercises
1. a. RS
b. RS
2. (continued)
d. Def. of segment
c. Given
d. Def. of bisector
e. Reflexive Prop.
of
bisector
e. Reflexive Prop.
of
f. SSS
f. AAS
g. CPCTC
5. VY; VT = VX (Ex. 3)
and UT = YX (Ex. 4),
so VU = VY by the
Subtr. Prop. of =.
6. Answers may vary.
Sample: VUY; s
opp. sides are .
7. x = 80; y = 40
2. a. KM
b. KM
c. By construction
3. VX; Conv. of the Isosc.
Thm.
4. UW; Conv. of the Isosc.
Thm.
4-5
8. x = 40; y = 70
9. x = 38; y = 4
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
10. x = 4 1 ; y = 60
2
19. a.
22. 140
11. x = 36; y = 36
23. 6
12. x = 92; y = 7
24. x = 60; y = 30
13. 64
1
14. 2 2
30, 30, 120
b. 5; 30, 60, 90, 120,
150
15. 42
16. 35
17. 150; 15
c. Check students’
work.
25. x = 64; y = 71
26. x = 30; y = 120
27. Two sides of a are
if and only if the s
opp. those sides
are .
20. 70
28. 80, 80, 20; 80, 50, 50
18. 24, 48, 72, 96, 120
21. 50
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
29. a. isosc. s
b. 900 ft; 1100 ft
c. The tower is the
bis. of the base of
each .
30. No; the can be
positioned in ways
such that the base is
not on the bottom.
31. 45; they are = and
have sum 90.
32. Answers may vary.
Sample: Corollary to
Thm. 4-3: Since XY
YZ, X
Z by Thm.
4-3. YZ ZX, so Y
X by Thm. 4-3 also.
By the Trans. Prop., Y
Z, so X
Y
Z. Corollary to Thm.
4-4: Since X
Z,
XY YZ by Thm. 4-4.
Y
X, so YZ ZX
by Thm. 4-4 also. By
the Trans. Prop. XY
ZX, so XY YZ ZX.
33. a. Given
b. A
c. Given
d. ABE
D
DCE
34. m = 36; n = 27
35. m = 60; n = 30
36. m = 20; n = 45
37. (0, 0), (4, 4), (–4, 0),
(0, –4), (8, 4), (4, 8)
38. (5, 0); (0, 5); (–5, 5);
(5, –5); (0, 10); (10, 0)
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
39. (5, 3); (2, 6); (2, 9);
(8, 3); (–1, 6); (5, 0)
40. a. 25
b. 40; 40; 100
c. Obtuse isosc. ;
2 of the s are
and one
obtuse.
is
41. AC CB and ACD
DCB are given. CD
CD by the Refl. Prop. of
, so ACD
BCD
by SAS. So AD DB
by CPCTC, and CD
bisects AB. Also ADC
BDC by CPCTC,
m ADC + m BDC =
180 by Add. Post., so
m ADC = m BDC =
90 by the Subst. Prop.
So CD is the bis.
of AB.
4-5
42. The bis. of the base
of an isosc. is the
bis. of the vertex ;
given isosc. ABC
with bis. CD, ADC
BDC and AD DB
by def. of bis. Since
CD CD by Refl.
Prop., ACD
BCD
by SAS. So ACD
BCD by CPCTC,
and CD bisects ACB.
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
43. a. 5
44. 0 < measure of base
< 45
b.
45. 45 < measure of base
< 90
49. [2] a. 60; since m PAB
= m PBA, and
m PAB +
m PBA = 120,
m PAB = 60.
b. 120; m APB =
60 so m PAB =
60. Since PAB
and QAB are
compl., m QAB
= 30. QAB
is isosc. so
m AQB = 120.
46. C
47. G
48. D
[1] one part correct
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
50. RC = GV; RC GV
by CPCTC since
RTC
GHV
by ASA.
51. AAS
52. SSS
53. 24 sides
4-5
Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Use the diagram for Exercises 1–3.
1. If mBAC = 38, find mC. 71
2. If mBAM = mCAM = 23, find
mBMA. 90
3. If mB = 3x and mBAC = 2x – 20,
find x. 25
4. Find the values of x and y.
x = 60
y=9
5. ABCDEF is a regular hexagon.
Find mBAC.
30
4-5
Congruence in Right Triangles
GEOMETRY LESSON 4-6
(For help, go to the Lessons 4-2 and 4-3.)
Tell whether the abbreviation identifies a congruence statement.
1. SSS
2. SAS
3. SSA
4. ASA
5. AAS
6. AAA
Can you conclude that the two triangles are congruent? Explain.
7.
8.
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
Solutions
1. SSS names the Side-Side-Side Theorem.
2. SAS names the Side-Angle-Side Theorem.
3. SSA does not name a theorem or postulate.
4. ASA names the Angle-Side-Angle Postulate.
5. AAS names the Angle-Angle-Side Theorem.
6. AAA does not name a congruence theorem or postulate.
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
Solutions (continued)
7. The side that they share is congruent to itself by the Reflexive Property
of Congruence. The two right angles are congruent to each other, and
two other corresponding sides are marked congruent. The two
triangles are congruent by SAS.
8. Two pairs of congruent sides are marked congruent. The included
angles are congruent because vertical angles are congruent. Thus, the
two triangles are congruent by SAS.
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
One student wrote “ CPA
MPA by SAS” for the diagram
below. Is the student correct? Explain.
The diagram shows the following congruent parts.
CA
MA
CPA
PA
MPA
PA
There are two pairs of congruent sides and one pair of
congruent angles, but the congruent angles are not
included between the corresponding congruent sides.
The triangles are not congruent by the SAS Postulate,
but they are congruent by the HL Theorem.
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
XYZ is isosceles. From vertex X, a perpendicular is drawn
to YZ, intersecting YZ at point M. Explain why XMY
XMZ.
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
Write a two–column proof.
Given: ABC and DCB are right angles, AC
Prove: ABC
DCB
Statements
DB
Reasons
1. ABC and DCB are
right angles.
2. ABC and DCB are
right triangles.
3. AC DB
4. BC CB
1. Given
5.
5. If the hypotenuse and a leg of one right
triangle are congruent to the hypotenuse
and a leg of another right triangle, then the
triangles are congruent. (HL Theorem).
ABC
DCB
2. Definition of a right triangle
3. Given
4. Reflexive Property of Congruence
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
Pages 219-223 Exercises
1.
ABC
DEF by HL. 4.
AEB
DCB by HL.
Both s are rt. s ,
Both s are rt. s .
AC DF, and CB FE.
AB BD and EB CB
by the def. of midpt.
2.
SPR
QRP by HL.
Both s are rt. s , SP
5.
T and Q are rt. s .
QR (given) and PR
PR by the Reflexive
6. RX RT or XV TV
Prop. of .
7. TY ER or RT YE
3.
LMP
OMN by HL.
Both s are rt. s
8. Right s are needed,
because vert. s are ;
either A and G or
LP NO, and LM
AQC and GJC.
OM.
4-6
9. BC
FA
10. RT
NQ
11. a. Given
b. Def. of rt.
c. Reflexive Prop.
of
d. Given
e. HL
Congruence in Right Triangles
GEOMETRY LESSON 4-6
12. a. suppl. s are rt.
b. Def. of rt.
c. Given
d. Reflexive Prop.
of
e. HL
s
15. Yes; PM PM and
PMW is a rt. since
JP || MW.
16. a. Given
b. Def. of
c. MLJ and
rt. s .
13. PS PT so S
T
d. Given
by the Isosc. Thm.
e. LJ LJ
PRS
PRT.
f. HL
PRS
PRT by
AAS.
17. a.
b.
c.
d.
e.
f.
KJL are
Given
IGH
Def. of rt.
I is the midpt. of HV.
Def. of midpt.
IGH
ITV
18. HL; each rt. has a
hyp. and side.
19. x = 3; y = 2
20. x = –1; y = 3
14. Yes; RS TU and
RT TV.
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
21. whether the 7-yd side
is the hyp. or a leg
22.
ABX
ABC
or SAS
BXC
ADX; HL
ADC; SSS
24.
23. a. Answers may vary.
Sample: You could
show that suppl. s
AXB and AXD are .
b.
DXC; HL
ABC
ADC by
SSS so BAC
DAC by CPCTC.
ABX
ADX by
SAS so AXB
AXD. AXB is
suppl. and to
AXD so they are
both rt. s .
4-6
25.
26.
27.
Congruence in Right Triangles
GEOMETRY LESSON 4-6
28. 1. EB DB; A and 29. 1. LO bisects MLN,
C are rt. s (Given)
OM LM, ON LN,
(Given)
2. BEA
BDC
are rt. s
2. M and N are rt. s
(Def. of rt. )
(Def. of )
3. B is the midpt. of
3. MLO
NLO
AC (Given)
(Def. of bis.)
4. AB BC
4. M
N
(Def. of midpt.)
(All rt. s are .)
5. BEA
BDC
5. LO LO (Reflexive
(HL)
Prop. of )
6. LMO
LNO
(AAS)
4-6
30. Answers may vary.
Sample: Measure 2
sides of the formed
by the amp. and the
platform’s corner.
Since the s will be
by HL or SAS, the s
are the same.
Congruence in Right Triangles
GEOMETRY LESSON 4-6
31. a.
b. slope of DG = –1;
slope of GF = –1;
slope of GE = 1
c. EGD and EGF
are rt. s .
d. DE = 26 ;
FE = 26
31. (continued)
e. EGD
EGF by
HL. Both s are rt. s ,
DE FE, and
EG EG.
32. An HA Thm. is the
same as AAS with AAS
corr. to the rt. , an
acute , and the hyp.
33. Since BE EA and
BE EC, AEB and
CEB are both rt. s .
AB BC because
ABC is equilateral,
and BE BE. AEB
CEB by HL.
34. No; AB CB because
AEB
CEB, but
doesn’t have to be
to AB or to CB.
35. A
36. H
4-6
Congruence in Right Triangles
GEOMETRY LESSON 4-6
37. D
38. [2] a.
b.
TFW
TGW
RFW and
RGW are rt. s .
41. BC || AD because each
slope = –1. BT BA,
BA AD because
product of slopes is –1.
42. If || lines then alt. int.
are .
s
[1] one part correct
39. isosceles
43. If two lines are ||, then
same-side int. s are
suppl.
40. equilateral
44. Vert.
s
are
.
45. If two lines are ||, then
corr. s are .
4-6
46. If two lines are ||, then
corr. s are .
47. If two lines are ||, then
same-side int. s are
suppl.
Congruence in Right Triangles
GEOMETRY LESSON 4-6
For Exercises 1 and 2, tell whether the HL Theorem can be used to prove
the triangles congruent. If so, explain. If not, write not possible.
1.
2.
Not
Yes; use the congruent
possible
hypotenuses and leg BC
to prove ABC
DCB
For Exercises 3 and 4, what additional information do you
need to prove the triangles congruent by the HL Theorem?
3. LMX
LOX
4. AMD
CNB
LM
LO
AM
CN
or
MD
4-6
NB
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
(For help, go to the Lessons 1-1 and 4-3.)
1. How many triangles will the next two figures in this pattern have?
2. Can you conclude that the triangles are congruent? Explain.
a. AZK and DRS
b. SDR and JTN
c. ZKA and NJT
4-7
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Solutions
1. For every new right triangle, segments connect the midpoint of the
hypotenuse with the midpoints of the legs of the right triangle, creating
two new triangles for every previous new triangle. The first figure has 1
triangle. The second has 1 + 2, or 3 triangles. The third has 3 + 4, or 7
triangles. The fourth will have 7 + 8, or 15 triangles. The fifth will have
15 + 16, or 31 triangles.
2. a. Two pairs of sides are congruent. The included angles are
congruent. Thus, the two triangles are congruent by SAS.
b. Two pairs of angles are congruent. One pair of sides is also
congruent, and, since it is opposite a pair of corresponding
congruent angles, the triangles are congruent by AAS.
c. Since AZK
Property of ,
DRS and SDR
ZKA
NJT.
4-7
JTN, by the Transitive
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Name the parts of their sides that
DFG and
Identify the overlapping triangles.
Parts of sides DG and EG are shared by
These parts are HG and FG, respectively.
4-7
DFG and
EHG.
EHG share.
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Write a Plan for Proof that does not use overlapping triangles.
Given: ZXW YWX, ZWX
Prove: ZW YX
YXW
Label point M where ZX intersects WY, as shown in the
diagram. ZW YX by CPCTC if ZWM
YXM.
You can prove these triangles congruent using ASA as follows:
Look at MWX. MW
Theorem.
MX by the Converse of the Isosceles Triangle
Look again at ZWM and YXM.  ZMW YMX because vertical
angles are congruent, MW MX, and by subtraction  ZWM YXM, so
ZWM
YXM by ASA.
4-7
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Write a paragraph proof.
Given: XW YZ, XWZ and YZW are right angles.
Prove: XPW
YPZ
Plan: XPW
YPZ by AAS if WXZ ZYW. These
angles are congruent by CPCTC if XWZ
YZW.
These triangles are congruent by SAS.
Proof: You are given XW YZ. Because XWZ and YZW are right angles,
XWZ YZW. WZ ZW, by the Reflexive Property of Congruence.
Therefore, XWZ
YZW by SAS. WXZ ZYW by CPCTC, and
XPW YPZ because vertical angles are congruent.
Therefore,
XPW
YPZ by AAS.
4-7
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Given: CA CE, BA DE
Write a two-column proof to show that CBE
Plan: CBE CDA by CPCTC if CBE
congruence holds by SAS if CB CD.
Reasons
Proof: Statements
1. BCE DCA
2. CA CE, BA DE
3. CA = CE, BA = DE
4. CA – BA = CE – DE
5. CA – BA = CB,
CE – DE = CD
6. CB = CD
7. CB CD
8. CBE
CDA
9. CBE CDA
CDA.
CDA. This
1. Reflexive Property of Congruence
2. Given
3. Congruent sides have equal measure.
4. Subtraction Property of Equality
5. Segment Addition Postulate
6. Substitution
7. Definition of congruence
8. SAS
9. CPCTC
4-7
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Pages 226-230 Exercises
1.
M
6.
9.
7.
10. a. Given
b. Reflexive Prop.
of
c. Given
2. DF
3. XY
4.
5.
d. AAS
e. CPCTC
8.
11.
4-7
LQP
PML; HL
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
12.
RST
UTS; SSS
16. AE DE if AEB
DEC by AAS. AB
13. QDA
UAD; SAS
DC and A
D since
they are corr. parts of
14. QPT
RUS; AAS
ABC and DCB,
which are by HL.
15. TD RO if TDI
ROE by AAS. TID 17. QET
QEU by
REO if TEI
SAS if QT QU. QT
RIE. TEI
RIE
and QU are corr. parts
by SSS.
of QTB and QUB
which are by ASA.
18.
19–22. Answers may
vary. Samples
are given.
19.
20.
4-7
ADC
EDG by
ASA if A
E. A
and E are corr.
parts in ADB and
EDF, which are
by SAS.
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
21. a.
23.
ACE
BCD by
ASA; AC BC, A
B (Given) C
C
(Reflexive Prop. of )
ACE
BCD (ASA)
25. m
m
m
m
m
24.
WYX
ZXY by HL;
WY YX, ZX YX, WX
ZY (Given) WYX
and ZXY are rt. s
(Def. of ) XY XY
(Reflexive Prop. of )
WYX
ZXY (HL)
26.
b.
22. a.
b.
1 = 56; m
3 = 34; m
5 = 22; m
7 = 34; m
9 = 112
ABC
FCG; ASA
27. a. Given
b. Reflexive Prop.
of
c. Given
d.
4-7
2 = 56;
4 = 90;
6 = 34;
8 = 68;
ETI
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
27. (continued)
e. IRE
28. a. Given
28. (continued)
i. DEC
b. Def. of
f. CPCTC
j. Vert.
s
are
c. Def. of rt.
g. Given
k. AAS
d. Given
h. All rt.
s
are
.
l. AE
e. BC
i.
BC
TDI
m. CPCTC
f. Reflexive
j.
DE
ROE
g. HL
k. CPCTC
h. CPCTC
4-7
.
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
29–30. Proofs may vary.
Samples are given.
29. It is given that 1
2 and 3
4.
Since QB QB by
the Reflexive Prop. of
, QTB
QUB
by ASA. So QT QU
by CPCTC. Since QE
QE by the
Reflexive Prop. of ,
then QET
QEU
by SAS.
30. 1. AD
ED (Given)
2. D is the midpt. of
BF. (Given)
30. (continued)
7. GDE
CDA
(Vert. s are .)
8.
3. FD DB (Def. of
midpt.)
4.
FDE
ADB
(Vert. s are .)
5.
FDE
(SAS)
BDA
6.
E
A (CPCTC)
4-7
ADC
(ASA)
31. a. AD
AB
AE
DE
BC;
DC;
EC;
EB
EDG
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
31. (continued)
32. 1. AC EC;
b. Use DB DB (refl.)
CB CD (Given)
and alt. int. s to
show ADB
2. C
C (Reflexive
CBD (ASA). AB
Prop. of )
DC and AD BC
(CPCTC). AEB
3. ACD
ECB
CED (ASA) and
(SAS)
AED
CEB
(ASA). Then AE
4. A
E (CPCTC)
EC and DE EB
(CPCTC).
33. PQ
RQ and PQT
RQT by Def. of
bisector. QT QT so
PQT
RQT by
SAS. P
R by
CPCTC. QT bisects
VQS so VQT
SQT and PQT
and RQT are both
rt. s . So VQP
SQR since they are
s.
compl. of
PQV
RQS by ASA so
QV QS by CPCTC.
34. C
35. F
4-7
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
36. A
38. [4] a. HL
37. [2] a.
HBC
HED
b. HB HE by
CPCTC if HBC
HED by ASA.
Since BDC
CED by AAS,
then DBC
CED by CPCTC
and CHB
DHE because
vertical s are .
[1] one part correct
b.
c. x = 30. In ADC m A + m ADC + m ACD
= 180. Substituting, 90 + x + x + x = 180.
Solving, x = 30.
d. 120; it is suppl. to a 60°
e. 6 m; DC = 2(AD)
4-7
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
38. (continued)
[3] 4 parts answered
correctly
45. y – 0 = –
40.
46–48. Eqs. may vary,
depending on pt.
chosen.
[2] 3 parts answered
correctly
46. y – 4 = 2(x – 1)
41.
[1] 2 parts answered
correctly
39. a. right
b.
1
(x – 0)
3
47. y + 5 =
1
2
42. y + 6 = (x – 2)
c. Reflexive
43. y – 5 = 1(x – 0)
d. HL
44. y – 6 = –2(x + 3)
4-7
5
(x – 3)
3
48. y + 3 = –
5
(x + 4)
6
Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
XY
1. Identify any common sides and
angles in AXY and BYX.
For Exercises 2 and 3, name a pair of
congruent overlapping triangles. State the
theorem or postulate that proves them congruent.
2.
3.
KSR
SAS
4. Plan a proof.
Given: AC BD, AD
Prove: XD XC
BC
MRS
GHI
ASA
IJG
XD XC by CPCTC if DXA
CXB.
This congruence holds by AAS if
BAD
ABC. Show BAD
ABC
by SSS.
4-7
Congruent Triangles
GEOMETRY CHAPTER 4
Page 236
1.
PAY
2.
ONE
APL
OSE
3. SAS
9. Answers may vary.
11. No; the lengths may
Sample: The corr. sides
be different.
of the two s may not
be .
12. 36
10. Answers may vary.
Sample:
4. HL
5. not possible
6. SSS
7. ASA
8. AAS
CE HD; CO HF;
EO DF; C
H;
E
D; O
F
4-A
13. AT || GS, so ATG
SGT because they
are alt. int. s . It is
given that AT GS,
and GT GT by the
Reflexive Prop. of ,
so GAT
TSG by
SAS.
Congruent Triangles
GEOMETRY CHAPTER 4
14. Since LN bisects
17. Answers may vary.
OLM and ONM,
Sample:
OLN
MLN and
ONL
MNL. LN
LN by the Reflexive
Prop. of , so OLN
MLN by ASA.
ABC
CDA
15. CFE
DEF; SSS
16.
TQS
TRA; SAS
4-A