1. 1 STANDARDS 1,2,3 END SHOW PARAGRAPH PROOF 1 PARAGRAPH PROOF 2 PARAGRAPH PROOF 3 PARAGRAPH PROOF 4 Two column proofs may also be written as paragraph proofs. All we need is to the statements and reasons with transition words, phrases or sentences, as may be seen in the following examples that convert two column proofs we did before to paragraph proofs. “glue” PARAGRAPH PROOF 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 2 Standard 1: Students demonstrate understanding by identifying and giving examples of undefined terms, axioms, theorems, and inductive and deductive reasoning. Standard 2: Students write geometric proofs, including proofs by contradiction. Standard 3: Students construct and judge the validity of a logical argument and give counterexamples to disprove a statement. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 Estándar 1: Los estudiantes demuestran entendimiento en identificar ejemplos de términos indefinidos, axiomas, teoremas, y razonamientos inductivos y deductivos. Standard 2: Los estudiantes escriben pruebas geométricas, incluyendo pruebas por contradicción. Standard 3: Los estudiantes construyen y juzgan la validéz de argumentos lógicos y dan contra ejemplos para desaprobar un estatuto. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 STANDARDS 1,2,3 Review the following two column proof and then convert it to paragraph proof: Two Column Proof: Statements Reasons (1) Given (2) (3) (4) L is midpoint of KM KL LM Definition of Midpoint LM AB Given KL AB of segments is transitive. Given: Prove: M LK B A L is midpoint of KM LM AB KL AB PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) Given (2) (3) (4) L is midpoint of KM KL LM Definition of Midpoint LM AB Given KL AB of segments is transitive. Given: Prove: M LK B A L is midpoint of KM LM AB KL AB The given states that L is midpoint of KM Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) Given (2) (3) (4) L is midpoint of KM KL LM Definition of Midpoint LM AB Given KL AB of segments is transitive. Given: Prove: M LK B A L is midpoint of KM LM AB KL AB The given states that L is midpoint of KM and therefore KL LM by definition of midpoint. Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) Given (2) (3) (4) L is midpoint of KM KL LM Definition of Midpoint LM AB Given KL AB of segments is transitive. Given: Prove: M LK B A L is midpoint of KM LM AB KL AB The given states that L is midpoint of KM and therefore KL LM by definition of midpoint. We also have that the given states LM AB Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) Given (2) (3) (4) L is midpoint of KM KL LM Definition of Midpoint LM AB Given KL AB of segments is transitive. Given: Prove: M LK B A L is midpoint of KM LM AB KL AB The given states that L is midpoint of KM and therefore KL LM by definition of midpoint. We also have that the given states LM AB. This allow us to conclude KL AB because of segments is transitive. Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 STANDARDS 1,2,3 Given: Prove: M LK B A L is midpoint of KM LM AB KL AB The given states that L is midpoint of KM and therefore KL LM by definition of midpoint. We also have that the given states LM AB. This allow us to conclude KL AB because of segments is transitive. PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 B C F D A E STANDARDS 1,2,3 Given: EFD is right Prove: AFB CFB and are complementary. Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right lines form 4 right s (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s B C F D A E Review the following two column proof and then convert it to paragraph proof: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 STANDARDS 1,2,3 Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E Convert to PARAGRAPH PROOF: From the given EFD is right Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s Lines form 4 right s PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 STANDARDS 1,2,3 Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E Convert to PARAGRAPH PROOF: From the given EFD is right and thus EC AD definition of lines by Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s Lines form 4 right s PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. 13 STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right Lines form 4 right s (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E Convert to PARAGRAPH PROOF: From the given EFD is right and thus EC AD definition of lines. by AFC is right We can also observe that because lines form 4 right s PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 14 STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right Lines form 4 right s (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E Convert to PARAGRAPH PROOF: From the given EFD is right and thus EC AD definition of lines. by AFC is right We can also observe that because lines form 4 right s and AFC= m 90° by definition of right s. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. 15 STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right Lines form 4 right s (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E Convert to PARAGRAPH PROOF: From the given EFD is right and thus EC AD definition of lines. by AFC is right We can also observe that because lines form 4 right s and AFC= m 90° by definition of right s. Now by addition postulate, it is that AFB + m AFC m CFB = m PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16. STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right Lines form 4 right s (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E Convert to PARAGRAPH PROOF: From the given EFD is right and thus EC AD definition of lines. by AFC is right We can also observe that because lines form 4 right s and AFC= m 90° by definition of right s. Now by addition postulate, it is that AFB + m AFC, m CFB = m where by substitution prop. of (=) results: AFB + m CFB = 90°, m PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17. STANDARDS 1,2,3 Two Column Proof: Statements Reasons (1) EFD is rightGiven (2) EC AD Definition of lines (3) AFC is right Lines form 4 right s (4) AFC= m 90° Definition of right s (5) AFB + m AFC m CFB = m (6) addition postulate AFB + m CFB = 90° m Substitution prop. of (=) AFB CFB and are complementary. (7) Definition of complementary s Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E Convert to PARAGRAPH PROOF: From the given EFD is right and thus EC AD definition of lines. by AFC is right We can also observe that because lines form 4 right s and AFC= m 90° by definition of right s. Now by addition postulate, it is that AFB + m AFC, m CFB = m where by substitution prop. of (=) results: AFB + m CFB = 90°, m and this last statement allow us to prove that AFB CFB and are complementary, by definition of complementary s. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18. 18 STANDARDS 1,2,3 Given: EFD is right Prove: AFB CFB and are complementary. B C F D A E From the given EFD is right and thus EC AD definition of lines. by AFC is right We can also observe that because lines form 4 right s and AFC= m 90° by definition of right s. Now by addition postulate, it is that AFB + m AFC, m CFB = m where by substitution prop. of (=) results: AFB + m CFB = 90°, m and this last statement allow us to prove that AFB CFB and are complementary, by definition of complementary s. PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19. 19 STANDARDS 1,2,3 Given: Prove: Two Column Proof: Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) B A C E D F G H CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m m BCD = 180° m 2( ) Substitution prop. of (=) Adding like terms B A C E D Review the following two column proof and then convert it to paragraph proof: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

22. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

23. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Now also from the given, we have that FGH ECA, Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

24. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Now also from the given, we have that FGH ECA, which implies FGH= m ECA m by definition of s. Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

25. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Now also from the given, we have that FGH ECA, which implies FGH= m ECA m by definition of s. This last statement allow us then by of s is transitive, to state that BCE= m FGH. m Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

26. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Now also from the given, we have that FGH ECA, which implies FGH= m ECA m by definition of s. This last statement allow us then by of s is transitive, to state that BCE= m FGH. m Now from the figure by BCE + m BCD = 180°, m ECA + m addition postulate, we have Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

27. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Now also from the given, we have that FGH ECA, which implies FGH= m ECA m by definition of s. This last statement allow us then by of s is transitive, to state that BCE= m FGH. m Now from the figure by BCE + m BCD = 180°, m ECA + m addition postulate, we have and this is FGH + m BCD = 180° m FGH + m by substitution prop. of (=). Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

28. STANDARDS B A C E D F G H Given: Prove: CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) FGH + m BCD = 180° m 2( ) Statements Reasons (1) (2) (3) (4) (5) (6) (7) (8) (9) CE bisectsBCA Given BCE ECA Definition of bisector BCE= m ECA m Definition of s FGH ECA Given FGH= m ECA m BCE= m FGH m Definition of s of s is transitive BCE + m BCD = 180° m ECA + m addition postulate FGH + m BCD = 180° m FGH + m Substitution prop. of (=) Adding like terms As stated in the given CE bisectsBCA, and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Now also from the given, we have that FGH ECA, which implies FGH= m ECA m by definition of s. This last statement allow us then by of s is transitive, to state that BCE= m FGH. m Now from the figure by BCE + m BCD = 180°, m ECA + m addition postulate, we have and this is FGH + m BCD = 180° m FGH + m by substitution prop. of (=). Adding like terms results FGH + m BCD = 180°. m 2( ) Convert to PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

29. 29 STANDARDS 1,2,3 Given: Prove: B A C E D F G H CE bisectsBCA FGH ECA FGH + m BCD = 180° m 2( ) As stated in the given and as consequence BCE ECA by definition of bisector, so that BCE= m ECA m by definition of s. Now also from the given, we have that FGH ECA, which implies FGH= m ECA m by definition of s. This last statement allow us then by of s is transitive, to state that BCE= m FGH. m Now from the figure by BCE + m BCD = 180°, m ECA + m addition postulate, we have and this is FGH + m BCD = 180° m FGH + m by substitution prop. of (=). Adding like terms results FGH + m BCD = 180°. m 2( ) PARAGRAPH PROOF: CE bisectsBCA, PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

30. 30 STANDARDS 1,2,3 Given: FBD is right Prove: ABF CBD and are complementary. Two Column Proof: Statements Reasons (1) (2) (3) (4) (5) (6) F E B C A D F E B C A D FBD is rightGiven FBD= m 90° Definition of right s FBD + m CBD = 180° m ABF + m addition postulate CBD = 180° m ABF + m 90° + ABF + m CBD = 90° m Substitution prop. of (=) Subtraction prop. of (=) ABF CBD and are complementary. Definition of complementary s Review the following two column proof and then convert it to paragraph proof: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

31. 31 STANDARDS 1,2,3 F E B C A D CBD = 180° m ABF + m 90° + Statements Reasons (1) (2) (3) (4) (5) (6) FBD is rightGiven FBD= m 90° Definition of right s FBD + m CBD = 180° m ABF + m addition postulate ABF + m CBD = 90° m Substitution prop. of (=) Subtraction prop. of (=) ABF CBD and are complementary. Definition of complementary s Given: FBD is right Prove: ABF CBD and are complementary. Convert to PARAGRAPH PROOF: The given is telling us that FBD is right, PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

32. 32 STANDARDS 1,2,3 F E B C A D CBD = 180° m ABF + m 90° + Statements Reasons (1) (2) (3) (4) (5) (6) FBD is rightGiven FBD= m 90° Definition of right s FBD + m CBD = 180° m ABF + m addition postulate ABF + m CBD = 90° m Substitution prop. of (=) Subtraction prop. of (=) ABF CBD and are complementary. Definition of complementary s Given: FBD is right Prove: ABF CBD and are complementary. Convert to PARAGRAPH PROOF: The given is telling us that FBD is right, then FBD= m 90° by definition of right s, PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

33. 33 STANDARDS 1,2,3 F E B C A D CBD = 180° m ABF + m 90° + Statements Reasons (1) (2) (3) (4) (5) (6) FBD is rightGiven FBD= m 90° Definition of right s FBD + m CBD = 180° m ABF + m addition postulate ABF + m CBD = 90° m Substitution prop. of (=) Subtraction prop. of (=) ABF CBD and are complementary. Definition of complementary s Given: FBD is right Prove: ABF CBD and are complementary. Convert to PARAGRAPH PROOF: The given is telling us that FBD is right, then FBD= m 90° by definition of right s, and from the figure: FBD + m CBD = 180° m ABF + m by addition postulate. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

34. 34 STANDARDS 1,2,3 F E B C A D CBD = 180° m ABF + m 90° + Statements Reasons (1) (2) (3) (4) (5) (6) FBD is rightGiven FBD= m 90° Definition of right s FBD + m CBD = 180° m ABF + m addition postulate ABF + m CBD = 90° m Substitution prop. of (=) Subtraction prop. of (=) ABF CBD and are complementary. Definition of complementary s Given: FBD is right Prove: ABF CBD and are complementary. Convert to PARAGRAPH PROOF: The given is telling us that FBD is right, then FBD= m 90° by definition of right s, and from the figure: FBD + m CBD = 180° m ABF + m by addition postulate. We can then have CBD = 180° m ABF + m 90° + using substitution prop. of (=). PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

35. 35 STANDARDS 1,2,3 F E B C A D CBD = 180° m ABF + m 90° + Statements Reasons (1) (2) (3) (4) (5) (6) FBD is rightGiven FBD= m 90° Definition of right s FBD + m CBD = 180° m ABF + m addition postulate ABF + m CBD = 90° m Substitution prop. of (=) Subtraction prop. of (=) ABF CBD and are complementary. Definition of complementary s Given: FBD is right Prove: ABF CBD and are complementary. Convert to PARAGRAPH PROOF: The given is telling us that FBD is right, then FBD= m 90° by definition of right s, and from the figure: FBD + m CBD = 180° m ABF + m by addition postulate. We can then have CBD = 180° m ABF + m 90° + using substitution prop. of (=). Now, if we subtract 90° from both sides applying subtraction prop. of (=), this results in ABF + m CBD = 90° m which is what we intended to prove, and that enable us to state that PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

36. 36 STANDARDS 1,2,3 F E B C A D CBD = 180° m ABF + m 90° + Statements Reasons (1) (2) (3) (4) (5) (6) FBD is rightGiven FBD= m 90° Definition of right s FBD + m CBD = 180° m ABF + m addition postulate ABF + m CBD = 90° m Substitution prop. of (=) Subtraction prop. of (=) ABF CBD and are complementary. Definition of complementary s Given: FBD is right Prove: ABF CBD and are complementary. Convert to PARAGRAPH PROOF: The given is telling us that FBD is right, then FBD= m 90° by definition of right s, and from the figure: FBD + m CBD = 180° m ABF + m by addition postulate. We can then have CBD = 180° m ABF + m 90° + using substitution prop. of (=). Now, if we subtract 90° from both sides applying subtraction prop. of (=), this results in ABF + m CBD = 90° m which is what we intended to prove, and that enable us to state that ABF CBD and are complementary, by definition of complementary s. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

37. 37 STANDARDS 1,2,3 Given: FBD is right Prove: ABF CBD and are complementary. F E B C A D The given is telling us that FBD is right, then FBD= m 90° by definition of right s, and from the figure: FBD + m CBD = 180° m ABF + m by addition postulate. We can then have CBD = 180° m ABF + m 90° + using substitution prop. of (=). Now, if we subtract 90° from both sides applying subtraction prop. of (=), this results in ABF + m CBD = 90° m which is what we intended to prove, and that enable us to state that ABF CBD and are complementary, by definition of complementary s. PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

38. 38 STANDARDS 1,2,3 Given: Prove: Two Column Proof: Statements Reasons (1) (2) (3) (4) (5) (6) (7) C A B H G D E F C A B H G D E F GE is a transversal AC and DF are GBC FEH and are supplementary. GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Review the following two column proof and then convert it to paragraph proof: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

39. 39 STANDARDS 1,2,3 C A B H G D E F Statements Reasons (1) (2) (3) (4) (5) (6) (7) GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Given: Prove: GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

40. 40 STANDARDS 1,2,3 C A B H G D E F Statements Reasons (1) (2) (3) (4) (5) (6) (7) GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Given: Prove: GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, therefore we have that CBE FEH because in lines cut by a transversal CORRESPONDING s are ; PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

41. 41 STANDARDS 1,2,3 C A B H G D E F Statements Reasons (1) (2) (3) (4) (5) (6) (7) GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Given: Prove: GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, therefore we have that CBE FEH because in lines cut by a transversal CORRESPONDING s are ; and this implies CBE= m FEH m by definition of s. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

42. 42 STANDARDS 1,2,3 C A B H G D E F Statements Reasons (1) (2) (3) (4) (5) (6) (7) GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Given: Prove: GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, therefore we have that CBE FEH because in lines cut by a transversal CORRESPONDING s are ; Now, we also have that GBC CBE and are a linear pair by def. of linear pair; and this implies CBE= m FEH m by definition of s. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

43. 43 STANDARDS 1,2,3 C A B H G D E F Statements Reasons (1) (2) (3) (4) (5) (6) (7) GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Given: Prove: GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, therefore we have that CBE FEH because in lines cut by a transversal CORRESPONDING s are ; Now, we also have that GBC CBE and are a linear pair by def. of linear pair; and this implies CBE= m FEH m by definition of s. so GBC + m CBE = 180° m because s in a linear pair are supplementary; PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

44. 44 STANDARDS 1,2,3 C A B H G D E F Statements Reasons (1) (2) (3) (4) (5) (6) (7) GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Given: Prove: GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, therefore we have that CBE FEH because in lines cut by a transversal CORRESPONDING s are ; Now, we also have that GBC CBE and are a linear pair by def. of linear pair; and this implies CBE= m FEH m by definition of s. so GBC + m CBE = 180° m because s in a linear pair are supplementary; this becomes GBC + m FEH = 180° m by substitution prop. of (=). PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

45. 45 STANDARDS 1,2,3 C A B H G D E F Statements Reasons (1) (2) (3) (4) (5) (6) (7) GE is a transversal AC and DF are Given GBC CBE and are a linear pair Definition of linear pair GBC + m CBE = 180° m s in a linear pair are supplementary CBE FEH In lines cut by a transversal CORRESPONDING s are Definition of s CBE= m FEH m GBC + m FEH = 180° m Substitution prop. of (=) GBC FEH and are supplementary. Definition of supplementary s Given: Prove: GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, therefore we have that CBE FEH because in lines cut by a transversal CORRESPONDING s are ; Now, we also have that GBC CBE and are a linear pair by def. of linear pair; and this implies CBE= m FEH m by definition of s. so GBC + m CBE = 180° m because s in a linear pair are supplementary; this becomes GBC + m FEH = 180° m by substitution prop. of (=). We can conclude GBC FEH and are supplementary, since they comply with definition of supplementary s. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

46. 46 STANDARDS 1,2,3 Given: Prove: C A B H G D E F GE is a transversal AC and DF are GBC FEH and are supplementary. The given states that AC and DF are and GE is a transversal, therefore we have that CBE FEH because in lines cut by a transversal CORRESPONDING s are ; Now, we also have that GBC CBE and are a linear pair by def. of linear pair; and this implies CBE= m FEH m by definition of s. so GBC + m CBE = 180° m because s in a linear pair are supplementary; this becomes GBC + m FEH = 180° m by substitution prop. of (=). We can conclude GBC FEH and are supplementary, since they comply with definition of supplementary s. PARAGRAPH PROOF: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved