warm up 1. if ∆ abc ∆ def , then a ? and bc ?
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EF. 17. Warm Up 1. If ∆ ABC ∆ DEF , then A ? and BC ? . 2. What is the distance between (3, 4) and (–1, 5)? 3. If 1 2, why is a||b ? 4. List the 4 theorems/postulates used to prove two triangles congruent:. D. - PowerPoint PPT PresentationTRANSCRIPT
Warm Up
1. If ∆ABC ∆DEF, then A ? and BC ? .
2. What is the distance between (3, 4) and (–1, 5)?
3. If 1 2, why is a||b?
4. List the 4 theorems/postulates used to prove two triangles congruent:
D EF
17
Converse of Alternate Interior Angles Theorem
SSS, SAS, ASA, AAS
Correcting Assignment #36(all but 17, 21)
20. 3 segments: 1 triangle3 angles: infinite triangles
Use CPCTC to prove parts of triangles are congruent.
Chapter 4.4 Using Corresponding Parts of
Congruent Triangles
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, and AAS use corresponding parts to prove triangles congruent. CPCTC uses congruent triangles to prove corresponding parts congruent. This is similar to the converse theorems in Chapter 3.
Remember!
Example 1: 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.
Check It Out! Example 1
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 2: Proving Corresponding Parts Congruent
Prove: XYW ZYW
Given: YW bisects XZ, XY YZ.
Z
Example 2 Continued
WY
ZW
Check It Out! Example 2
Prove: PQ PS
Given: PR bisects QPS and QRS.
Check It Out! Example 2 Continued
PR bisects QPS
and QRS
QRP SRP
QPR SPR
Given Def. of bisector
RP PR
Reflex. Prop. of
∆PQR ∆PSR
PQ PS
ASA
CPCTC
Example 3: 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 3 Continued
Assignment #37: Pages 246-248
Foundation: 6, 7
Core: 9, 10
Review: 27-32
Check It Out! Example 3
Prove: KL || MN
Given: J is the midpoint of KM and NL.
Check It Out! Example 3 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
Lesson Quiz: Part I
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
Lesson Quiz: Part I Continued
Lesson Quiz: Part II
2. Given: X is the midpoint of AC . 1 2
Prove: X is the midpoint of BD.
Lesson Quiz: Part II Continued
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