1 / 19

CPCTC

CPCTC. B. C. Given: E is the midpoint of BD <B = <D Prove: AB = CD. E. <1. <2. A. D. Statement Reason E is the midpoint of BD given <B = <D given. B. C. Given: E is the midpoint of BD <B = <D Prove: AB = CD. E. <1. <2. A. D. Statement Reason

pcoats
Download Presentation

CPCTC

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CPCTC

  2. B C Given: E is the midpoint of BD <B = <D Prove: AB = CD E <1 <2 A D Statement Reason E is the midpoint of BD given <B = <D given

  3. B C Given: E is the midpoint of BD <B = <D Prove: AB = CD E <1 <2 A D Statement Reason E is the midpoint of BD given <B = <D given BE = DE definition of a midpoint <1 = <2 vertical angle theorem ∆ABC = ∆CDE ASA AB = CD CPCTC

  4. B C Given: E is the midpoint of BD <A = <C Prove: <B = <D E <1 <2 A D Statement Reason E is the midpoint of BD given <A = <C given

  5. B C Given: E is the midpoint of BD <A = <C Prove: <B = <D E <1 <2 A D Statement Reason E is the midpoint of BD given <A = <C given BE = DE definition of a midpoint <1 = <2 vertical angle theorem ∆ABC = ∆CDE AAS <B = <D CPCTC

  6. B C Given: AB // DC AB = DC Prove: BE = DE (using ASA) E <2 <5 <3 <4 <1 <6 A D Statement Reason AB // DC given AB = DC given

  7. B C Given: AB // DC AB = DC Prove: BE = DE (using ASA) E <2 <5 <3 <4 <1 <6 A D Statement Reason AB // DC given AB = DC given <2 = <6 alternate interior angle theorem <1 = <5 alternate interior angle theorem ∆ABC = ∆CDE ASA BE = DE CPCTC

  8. B C Given: AB // DC AB = DC Prove: AE = CE (using AAS) E <2 <5 <3 <4 <1 <6 A D Statement Reason AB // DC given AB = DC given

  9. B C Given: AB // DC AB = DC Prove: AE = CE (using AAS) E <2 <5 <3 <4 <1 <6 A D Statement Reason AB // DC given AB = DC given BE = DE definition of a midpoint <1 = <5 alternate interior angle theorem <3 = <4 vertical angle theorem ∆ABC = ∆CDE AAS AE = CE CPCTC

  10. B Given: AB = AD BC = DC Prove: <1 = <2 <1 <3 A C <2 <4 D Statement Reason AB = AD given BC = DC given

  11. B Given: AB = AD BC = DC Prove: <1 = <2 <1 <3 A C <2 <4 D Statement Reason AB = AD given BC = DC given AC = AC reflexive property ∆ABC = ∆ADC SSS <1 = <2 CPCTC

  12. B Given: AC bisects <A AC bisects <C Prove: <B = <D <1 <3 A C <2 <4 D Statement Reason AC bisects <A given AC bisects <C given

  13. B Given: AC bisects <A AC bisects <C Prove: <B = <D <1 <3 A C <2 <4 D Statement Reason AC bisects <A given AC bisects <C given <1 = <2 definition of an angle bisector <3 = <4 definition of an angle bisector AC = AC reflexive property ∆ABC = ∆ADC ASA <B = <D CPCTC

  14. B Given: D is the midpoint of AC AB = CB Prove: <1 = <2 <1 <2 <3 <4 A C D Statement Reason D is the midpoint of AC given AB = CB given

  15. B Given: D is the midpoint of AC AB = CB Prove: <1 = <2 <1 <2 <3 <4 A C D Statement Reason D is the midpoint of AC given AB = CB given AD = CD definition of a midpoint BD = BD reflexive property ∆ABD = ∆CBD SSS <1 = <2 CPCTC

  16. B Given: <4 = 90° D is the midpoint of AC Prove: <A = <C <1 <2 <3 <4 A C D Statement Reason <4 = 90° given D is the midpoint of AC given

  17. B Given: <4 = 90° D is the midpoint of AC Prove: <A = <C <1 <2 <3 <4 A C D Statement Reason <4 = 90° given D is the midpoint of AC given <3 + <4 = 180 linear pair postulate <3 + 90 = 180 substitution <3 = 90 subtraction <3 = <4 substitution AD = CD definition of a midpoint BD = BD reflexive property ∆ABD = ∆CBD SAS <A = <C CPCTC

  18. B Given: AC bisects <A AC bisects <C Prove: <B = <D (w/o using ∆ congruence) <1 <3 A C <2 <4 D Statement Reason AC bisects <A given AC bisects <C given

  19. B Given: AC bisects <A AC bisects <C Prove: <B = <D (w/o using ∆ congruence) <1 <3 A C <2 <4 D Statement Reason AC bisects <A given AC bisects <C given <1 = <2 definition of angle bisector <3 = <4 definition of angle bisector <1 + <3 + <B= 180 triangle sum theorem <2 + <4 + <D = 180 triangle sum theorem <1 + <3 + <B = <2 + <4 + <D substitution <2 + <4 + <B = <2 + <4 + <D substitution <B = <D subtraction property

More Related