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Angle Relationships in Triangles
Holt Geometry
Lesson PresentationLesson Presentation
Holt 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.
Interior
Exterior
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.
Interior
Exterior
4 is an exterior angle.
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.
Interior
Exterior
3 is an interior angle.
4 is an exterior angle.
The remote interior angles of 4 are 1 and 2.
Find mB.
Example 3: Applying the Exterior Angle Theorem
mA + mB = mBCD Ext. Thm.
15 + 2x + 3 = 5x – 60 Substitute 15 for mA, 2x + 3 for mB, and 5x – 60 for mBCD.
2x + 18 = 5x – 60 Simplify.
78 = 3xSubtract 2x and add 60 to both sides.
26 = x Divide by 3.
mB = 2x + 3 = 2(26) + 3 = 55°
Find mACD.
Example 3
mACD = mA + mB Ext. Thm.
6z – 9 = 2z + 1 + 90 Substitute 6z – 9 for mACD, 2z + 1 for mA, and 90 for mB.
6z – 9 = 2z + 91 Simplify.
4z = 100Subtract 2z and add 9 to both sides.
z = 25 Divide by 4.
mACD = 6z – 9 = 6(25) – 9 = 141°
Example 4
Find mP and mT.
P T
mP = mT
2x2 = 4x2 – 32
–2x2 = –32
x2 = 16
So mP = 2x2 = 2(16) = 32°.
Since mP = mT, mT = 32°.
Third s Thm.
Def. of s.
Substitute 2x2 for mP and 4x2 – 32 for mT.
Subtract 4x2 from both sides.
Divide both sides by -2.
Practice
1. Find mABD. 2. Find mN and mP.
124°75°; 75°
Congruent Triangles
Holt Geometry
Lesson PresentationLesson Presentation
Holt 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.
When you write a statement such as ABC DEF, you are also stating which parts are congruent.
Helpful Hint
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
If polygon LMNP polygon EFGH, identify all pairs of corresponding congruent parts.
Example 1
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.
BCA BCD
mBCA = mBCD
(2x – 16)° = 90°
2x = 106
x = 53
Def. of lines.
Rt. Thm.
Def. of s
Substitute values for mBCA and mBCD.
Add 16 to both sides.
Divide both sides by 2.
Example 2B
Given: ∆ABC ∆DBC.
Find mDBC.
mABC + mBCA + mA = 180°
mABC + 90 + 49.3 = 180
mABC + 139.3 = 180
mABC = 40.7
DBC ABC
mDBC = mABC
∆ Sum Thm.
Substitute values for mBCA and mA.Simplify.
Subtract 139.3 from both sides.
Corr. s of ∆s are .
Def. of s.
mDBC 40.7° Trans. Prop. of =
Given: ∆ABC ∆DEF
Example 3
Find the value of x.
2x – 2 = 6
2x = 8
x = 4
Corr. sides of ∆s are .
Add 2 to both sides.
Divide both sides by 2.
AB DE
Substitute values for AB and DE.
AB = DE Def. of parts.
Given: ∆ABC ∆DEF
Example 4
Find mF.
mEFD + mDEF + mFDE = 180°
mEFD + 53 + 90 = 180
mF + 143 = 180
mF = 37°
ABC DEF
mABC = mDEF
∆ Sum Thm.
Substitute values for mDEF and mFDE.
Simplify.
Subtract 143 from both sides.
Corr. s of ∆ are .
Def. of s.
mDEF = 53° Transitive Prop. of =.
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
9. Given
7. Reflex. Prop. of
8. Third s Thm.8. X Z
10. Def. of ∆10. ∆XYW ∆ZYW
6. Def. of mdpt.
5. Given
4. XYW ZYW 4. Def. of bisector
9. XY YZ
7. YW YW
6. XW ZW
3. Given3. YW bisects XYZ
2. Rt. Thm.2. YWX YWZ
1. Given1. YWX and YWZ are rt. s.
5. W is mdpt. of XZ
Example
Given: AD bisects BE.
BE bisects AD. AB DE, A D Prove: ∆ABC ∆DEC
6. Def. of bisector
7. Def. of ∆s7. ∆ABC ∆DEC
5. Given
3. ABC DEC
4. Given
2. BCA DCE
3. Third s Thm.
2. Vertical s are .
1. Given1. A D
4. AB DE
Statements Reasons
BE bisects AD
5. AD bisects BE,
6. BC EC, AC DC
Triangle Congruence: SSS and SAS
Holt Geometry
Lesson PresentationLesson Presentation
Holt 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.
The letters SAS are written in that order because the congruent angles must be between pairs of congruent corresponding sides.
Caution
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 ADProve: ∆ABD ∆CDB
ReasonsStatements
5. SAS Steps 3, 2, 45. ∆ABD ∆ CDB
4. Reflex. Prop. of
3. Given
2. Alt. Int. s Thm.2. CBD ADB
1. Given1. BC || AD
3. BC AD
4. BD BD
Check It Out! Example 4
Given: QP bisects RQS. QR QS
Prove: ∆RQP ∆SQP
ReasonsStatements
5. SAS Steps 1, 3, 45. ∆RQP ∆SQP
4. Reflex. Prop. of
1. Given
3. Def. of bisector3. RQP SQP
2. Given2. QP bisects RQS
1. QR QS
4. QP QP
Practice
Given: PN bisects MO, PN MO
Prove: ∆MNP ∆ONP
1. Given2. Def. of bisect3. Reflex. Prop. of 4. Given5. Def. of 6. Rt. Thm.7. SAS Steps 2, 6, 3
1. PN bisects MO2. MN ON3. PN PN4. PN MO 5. PNM and PNO are rt. s6. PNM PNO
7. ∆MNP ∆ONP
Reasons Statements
Triangle Congruence: ASA, AAS, and HL
Holt Geometry
Lesson PresentationLesson Presentation
Holt 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 VYProve: 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.
ASAHL
SAS or SSS
Practice
4. Given: FAB GED, ABC DCE, AC EC
Prove: ABC EDC
Lesson Quiz: Part II Continued
5. ASA Steps 3,45. ABC EDC
4. Given4. ACB DCE; AC EC
3. Supp. Thm.3. BAC DEC
2. Def. of supp. s2. BAC is a supp. of FAB; DEC is a supp. of GED.
1. Given1. FAB GED
ReasonsStatements
Triangle Congruence: CPCTC
Holt Geometry
Lesson PresentationLesson Presentation
Holt 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.
SSS, SAS, ASA, AAS, and HL use corresponding parts to prove triangles congruent. CPCTC uses congruent triangles to prove corresponding parts congruent.
Remember!
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
Prove: XYW ZYW
Given: YW bisects XZ, XY YZ.
Z
Example
Prove: PQ PS
Given: PR bisects QPS and QRS.
Example: Using CPCTC in a Proof
Prove: MN || OP
Given: NO || MP, N P
5. CPCTC5. NMO POM
6. Conv. Of Alt. Int. s Thm.
4. AAS4. ∆MNO ∆OPM
3. Reflex. Prop. of
2. Alt. Int. s Thm.2. NOM PMO
1. Given
ReasonsStatements
3. MO MO
6. MN || OP
1. N P; NO || MP
Example Continued
Example
Prove: KL || MN
Given: J is the midpoint of KM and NL.
Example Continued
5. CPCTC5. LKJ NMJ
6. Conv. Of Alt. Int. s Thm.
4. SAS Steps 2, 34. ∆KJL ∆MJN
3. Vert. s Thm.3. KJL MJN
2. Def. of mdpt.
1. Given
ReasonsStatements
6. KL || MN
1. J is the midpoint of KM and NL.
2. KJ MJ, NJ LJ
Practice
1. Given: Isosceles ∆PQR, base QR, PA PB
Prove: AR BQ
4. Reflex. Prop. of 4. P P
5. SAS Steps 2, 4, 35. ∆QPB ∆RPA
6. CPCTC6. AR = BQ
3. Given3. PA = PB
2. Def. of Isosc. ∆2. PQ = PR
1. Isosc. ∆PQR, base QR
Statements
1. Given
Reasons
Practice Solution
Practice
2. Given: X is the midpoint of AC . 1 2
Prove: X is the midpoint of BD.
Practice 2 Solution
6. CPCTC
7. Def. of 7. DX = BX
5. ASA Steps 1, 4, 55. ∆AXD ∆CXB
8. Def. of mdpt.8. X is mdpt. of BD.
4. Vert. s Thm.4. AXD CXB
3. Def of 3. AX CX
2. Def. of mdpt.2. AX = CX
1. Given1. X is mdpt. of AC. 1 2
ReasonsStatements
6. DX BX
