# 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.

## Presentation on theme: "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."— Presentation transcript:

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 D EF 17 Converse of Alternate Interior Angles Theorem SSS, SAS, ASA, AAS

Correcting Assignment #36 (all but 17, 21) 20. 3 segments: 1 triangle 3 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. CPCTC 5. NMO  POM 6. Conv. Of Alt. Int. s Thm. 4. AAS 4. ∆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. CPCTC 5. LKJ  NMJ 6. Conv. Of Alt. Int. s Thm. 4. SAS Steps 2, 3 4. ∆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, 3 5. ∆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, 5 5. ∆ 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. Given 1. X is mdpt. of AC. 1  2 ReasonsStatements 6. DX  BX

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