Angle Relationships in Triangles
Lesson Presentation
Holt
HoltGeometry
McDougal Geometry
Review Topics
•
•
•
•
•
Vertical Angles
Parallel Lines cut by a Transversal
Linear Angles (Straight Angle)
Complementary and Supplementary Angles
Reflexive, Symmetric and Transitive
Properties
• Congruence
• Definition of Bisector and Midpoint
An auxiliary line is a line that is added to a
figure to aid in a proof.
An auxiliary
line used in the
Triangle Sum
Theorem
A corollary is a theorem whose proof follows
directly from another theorem. Here are two
corollaries to the Triangle Sum Theorem.
The interior is the set of all points inside the
figure. The exterior is the set of all points
outside the figure.
Exterior
Interior
An interior angle is formed by two sides of a
triangle. An exterior angle is formed by one
side of the triangle and extension of an adjacent
side.
4 is an exterior angle.
Exterior
Interior
3 is an interior angle.
Each exterior angle has two remote interior angles. A
remote interior angle is an interior angle that is not
adjacent to the exterior angle.
Exterior
Interior
3 is an interior angle.
4 is an exterior angle.
The remote interior
angles of 4 are 1
and 2.
Example 3: Applying the Exterior Angle Theorem
Find mB.
mA + mB = mBCD
Ext.  Thm.
15 + 2x + 3 = 5x – 60
Substitute 15 for mA, 2x + 3 for
mB, and 5x – 60 for mBCD.
2x + 18 = 5x – 60
78 = 3x
Simplify.
Subtract 2x and add 60 to
both sides.
Divide by 3.
26 = x
mB = 2x + 3 = 2(26) + 3 = 55°
Example 3
Find mACD.
mACD = mA + mB
Ext.  Thm.
6z – 9 = 2z + 1 + 90
Substitute 6z – 9 for mACD,
2z + 1 for mA, and 90 for mB.
6z – 9 = 2z + 91
Simplify.
4z = 100
Subtract 2z and add 9 to both
sides.
Divide by 4.
z = 25
mACD = 6z – 9 = 6(25) – 9 = 141°
Example 4
Find mP and mT.
P  T
mP = mT
Third s Thm.
Def. of  s.
2x2 = 4x2 – 32 Substitute 2x2 for mP and 4x2 – 32 for mT.
–2x2 = –32
x2 = 16
Subtract 4x2 from both sides.
Divide both sides by -2.
So mP = 2x2 = 2(16) = 32°.
Since mP = mT, mT = 32°.
Practice
1. Find mABD.
124°
2. Find mN and mP.
75°; 75°
Congruent Triangles
Lesson Presentation
Holt
HoltGeometry
McDougal Geometry
Geometric figures are congruent if they are the
same size and shape. Corresponding angles
and corresponding sides are in the same
position in polygons with an equal number of
sides.
Two polygons are congruent polygons if and
only if their corresponding sides are congruent.
Thus triangles that are the same size and shape
are congruent.
Helpful Hint
When you write a statement such as
ABC  DEF, you are also stating
which parts are congruent.
EXAMPLE
Given: ∆PQR  ∆STW
Identify all pairs of corresponding congruent parts.
Angles: P  S, Q  T, R  W
Sides: PQ  ST, QR  TW, PR  SW
Example 1
If polygon LMNP  polygon EFGH, identify all
pairs of corresponding congruent parts.
Angles: L  E, M  F, N  G, P  H
Sides: LM  EF, MN  FG, NP  GH, LP  EH
Example 2
Given: ∆ABC  ∆DBC.
Find the value of x.
BCA and BCD are rt. s.
Def. of  lines.
BCA  BCD
Rt.   Thm.
mBCA = mBCD
Def. of  s
(2x – 16)° = 90°
2x = 106
x = 53
Substitute values for mBCA and
mBCD.
Add 16 to both sides.
Divide both sides by 2.
Example 2B
Given: ∆ABC  ∆DBC.
Find mDBC.
mABC + mBCA + mA = 180° ∆ Sum Thm.
Substitute values for mBCA and
mABC + 90 + 49.3 = 180
mA.
mABC + 139.3 = 180 Simplify.
mABC = 40.7
DBC  ABC
Subtract 139.3 from both
sides.
Corr. s of  ∆s are  .
mDBC = mABC Def. of  s.
mDBC  40.7°
Trans. Prop. of =
Example 3
Given: ∆ABC  ∆DEF
Find the value of x.
AB  DE
Corr. sides of  ∆s are .
AB = DE
Def. of  parts.
2x – 2 = 6
2x = 8
x=4
Substitute values for AB and DE.
Add 2 to both sides.
Divide both sides by 2.
Example 4
Given: ∆ABC  ∆DEF
Find mF.
mEFD + mDEF + mFDE = 180°
ABC  DEF
∆ Sum Thm.
Corr. s of  ∆ are .
mABC = mDEF
Def. of  s.
mDEF = 53°
Transitive Prop. of =.
mEFD + 53 + 90 = 180
mF + 143 = 180
mF = 37°
Substitute values for mDEF
and mFDE.
Simplify.
Subtract 143 from both sides.
Example : Proving Triangles Congruent
Given: YWX and YWZ are right angles.
YW bisects XYZ. W is the midpoint of XZ. XY  YZ.
Prove: ∆XYW  ∆ZYW
Statements
Reasons
1. YWX and YWZ are rt. s.
1. Given
2. YWX  YWZ
2. Rt.   Thm.
3. YW bisects XYZ
3. Given
4. XYW  ZYW
4. Def. of bisector
5. W is mdpt. of XZ
5. Given
6. XW  ZW
6. Def. of mdpt.
7. YW  YW
7. Reflex. Prop. of 
8. X  Z
8. Third s Thm.
9. XY  YZ
9. Given
10. ∆XYW  ∆ZYW
10. Def. of  ∆
Example
Given: AD bisects BE.
BE bisects AD.
AB  DE, A  D
Prove: ∆ABC  ∆DEC
Statements
Reasons
1. A  D
1. Given
2. BCA  DCE
2. Vertical s are .
3. ABC  DEC
3. Third s Thm.
4. AB  DE
4. Given
5. AD bisects BE,
5. Given
BE bisects AD
6. BC  EC, AC  DC
6. Def. of bisector
7. ∆ABC  ∆DEC
7. Def. of  ∆s
Triangle Congruence: SSS and SAS
Lesson Presentation
Holt
HoltGeometry
McDougal Geometry
For example, you only need to know that
two triangles have three pairs of congruent
corresponding sides. This can be expressed
as the following postulate.
Example 1: Using SSS to Prove Triangle Congruence
Use SSS to explain why ∆ABC  ∆DBC.
It is given that AC  DC and that AB  DB. By the
Reflexive Property of Congruence, BC  BC.
Therefore ∆ABC  ∆DBC by SSS.
An included angle is an angle formed
by two adjacent sides of a polygon.
B is the included angle between sides
AB and BC.
Caution
The letters SAS are written in that order
because the congruent angles must be
between pairs of congruent corresponding
sides.
Example: Engineering Application
The diagram shows part of
the support structure for a
tower. Use SAS to explain
why ∆XYZ  ∆VWZ.
It is given that XZ  VZ and that YZ  WZ.
By the Vertical s Theorem. XZY  VZW.
Therefore ∆XYZ  ∆VWZ by SAS.
Example
Use SAS to explain why
∆ABC  ∆DBC.
It is given that BA  BD and ABC  DBC.
By the Reflexive Property of , BC  BC.
So ∆ABC  ∆DBC by SAS.
Example 4: Proving Triangles Congruent
Given: BC ║ AD, BC  AD
Prove: ∆ABD  ∆CDB
Statements
Reasons
1. BC || AD
1. Given
2. CBD  ADB
2. Alt. Int. s Thm.
3. BC  AD
3. Given
4. BD  BD
4. Reflex. Prop. of 
5. ∆ABD  ∆ CDB
5. SAS Steps 3, 2, 4
Check It Out! Example 4
Given: QP bisects RQS. QR  QS
Prove: ∆RQP  ∆SQP
Statements
Reasons
1. QR  QS
1. Given
2. QP bisects RQS
2. Given
3. RQP  SQP
3. Def. of bisector
4. QP  QP
4. Reflex. Prop. of 
5. ∆RQP  ∆SQP
5. SAS Steps 1, 3, 4
Practice
Given: PN bisects MO, PN  MO
Prove: ∆MNP  ∆ONP
Statements
Reasons
1.
2.
3.
4.
5.
6.
1.
2.
3.
4.
5.
6.
7.
PN bisects MO
MN  ON
PN  PN
PN  MO
PNM and PNO are rt. s
PNM  PNO
7. ∆MNP  ∆ONP
Given
Def. of bisect
Reflex. Prop. of 
Given
Def. of 
Rt.   Thm.
SAS Steps 2, 6, 3
Triangle Congruence: ASA, AAS, and HL
Lesson Presentation
Holt
HoltGeometry
McDougal Geometry
An included side is the common side
of two consecutive angles in a polygon.
The following postulate uses the idea of
an included side.
Example: Applying ASA Congruence
Determine if you can use ASA to prove the
triangles congruent. Explain.
Two congruent angle pairs are give, but the included
sides are not given as congruent. Therefore ASA
cannot be used to prove the triangles congruent.
Example
Determine if you can use ASA to
prove NKL  LMN. Explain.
By the Alternate Interior Angles Theorem. KLN  MNL.
NL  LN by the Reflexive Property. No other congruence
relationships can be determined, so ASA cannot be
applied.
You can use the Third Angles Theorem to prove
another congruence relationship based on ASA. This
theorem is Angle-Angle-Side (AAS).
Example: Using AAS to Prove Triangles Congruent
Use AAS to prove the triangles congruent.
Given: X  V, YZW  YWZ, XY  VY
Prove:  XYZ  VYW
Example
Use AAS to prove the triangles congruent.
Given: JL bisects KLM, K  M
Prove: JKL  JML
Example: Applying HL Congruence
Determine if you can use the HL Congruence
Theorem to prove the triangles congruent. If
not, tell what else you need to know.
According to the diagram,
the triangles are right
triangles that share one
leg.
It is given that the
hypotenuses are
congruent, therefore the
triangles are congruent by
HL.
Example: Applying HL Congruence
This conclusion cannot be proved by HL. According
to the diagram, the triangles are right triangles and
one pair of legs is congruent. You do not know that
one hypotenuse is congruent to the other.
Example
Determine if you can use
the HL Congruence Theorem
to prove ABC  DCB. If
not, tell what else you need
to know.
Yes; it is given that AC  DB. BC  CB by the
Reflexive Property of Congruence. Since ABC
and DCB are right angles, ABC and DCB are
right triangles. ABC  DCB by HL.
Practice
Identify the postulate or theorem that proves
the triangles congruent.
ASA
HL
SAS or SSS
Practice
4. Given: FAB  GED, ABC   DCE, AC  EC
Prove: ABC  EDC
Lesson Quiz: Part II Continued
Statements
Reasons
1. FAB  GED
1. Given
2. BAC is a supp. of FAB;
DEC is a supp. of GED.
2. Def. of supp. s
3. BAC  DEC
3.  Supp. Thm.
4. ACB  DCE; AC  EC
4. Given
5. ABC  EDC
5. ASA Steps 3,4
Triangle Congruence: CPCTC
Lesson Presentation
Holt
HoltGeometry
McDougal Geometry
CPCTC is an abbreviation for the phrase
“Corresponding Parts of Congruent
Triangles are Congruent.” It can be used
as a justification in a proof after you have
proven two triangles congruent.
Remember!
SSS, SAS, ASA, AAS, and HL use
corresponding parts to prove triangles
congruent. CPCTC uses congruent
triangles to prove corresponding parts
congruent.
Example: Engineering Application
A and B are on the edges
of a ravine. What is AB?
Example
A landscape architect sets
up the triangles shown in
the figure to find the
distance JK across a pond.
What is JK?
Example: Engineering Application
A and B are on the edges
of a ravine. What is AB?
One angle pair is congruent,
because they are vertical
angles. Two pairs of sides
are congruent, because their
lengths are equal.
Therefore the two triangles are congruent by
SAS. By CPCTC, the third side pair is congruent,
so AB = 18 mi.
Example
A landscape architect sets
up the triangles shown in
the figure to find the
distance JK across a pond.
What is JK?
One angle pair is congruent,
because they are vertical
angles.
Two pairs of sides are congruent, because their
lengths are equal. Therefore the two triangles are
congruent by SAS. By CPCTC, the third side pair is
congruent, so JK = 41 ft.
Example: Proving Corresponding Parts Congruent
Given: YW bisects XZ, XY  YZ.
Prove: XYW  ZYW
Z
Example
Given: PR bisects QPS and QRS.
Prove: PQ  PS
Example: Using CPCTC in a Proof
Given: NO || MP, N  P
Prove: MN || OP
Example Continued
Statements
Reasons
1. N  P; NO || MP
1. Given
2. NOM  PMO
2. Alt. Int. s Thm.
3. MO  MO
3. Reflex. Prop. of 
4. ∆MNO  ∆OPM
4. AAS
5. NMO  POM
5. CPCTC
6. MN || OP
6. Conv. Of Alt. Int. s Thm.
Example
Given: J is the midpoint of KM and NL.
Prove: KL || MN
Example Continued
Statements
Reasons
1. J is the midpoint of KM
and NL.
1. Given
2. KJ  MJ, NJ  LJ
2. Def. of mdpt.
3. KJL  MJN
3. Vert. s Thm.
4. ∆KJL  ∆MJN
4. SAS Steps 2, 3
5. LKJ  NMJ
5. CPCTC
6. KL || MN
6. Conv. Of Alt. Int. s
Thm.
Practice
1. Given: Isosceles ∆PQR, base QR, PA  PB
Prove: AR  BQ
Practice Solution
Statements
Reasons
1. Isosc. ∆PQR, base QR
1. Given
2. PQ = PR
2. Def. of Isosc. ∆
3. PA = PB
3. Given
4. P  P
4. Reflex. Prop. of 
5. ∆QPB  ∆RPA
5. SAS Steps 2, 4, 3
6. AR = BQ
6. CPCTC
Practice
2. Given: X is the midpoint of AC . 1  2
Prove: X is the midpoint of BD.
Practice 2 Solution
Statements
Reasons
1. X is mdpt. of AC. 1  2
1. Given
2. AX = CX
2. Def. of mdpt.
3. AX  CX
3. Def of 
4. AXD  CXB
4. Vert. s Thm.
5. ∆AXD  ∆CXB
5. ASA Steps 1, 4, 5
6. DX  BX
6. CPCTC
7. DX = BX
7. Def. of 
8. X is mdpt. of BD.
8. Def. of mdpt.