Indirect Proof: That the translation of by vector will lie on. E The way an indirect proof works is to assume the opposite and prove that it creates a contradiction so that the opposite must be true. So let us assume that point the image of translated by L, E vector does not lie on. The contradiction comes when we prove that ∆DBE ∆LEG but then must equal but it doesn’t…thus the EG EC contradiction verifies that must be on. L = because is the midpoint and = because translation vector thus = by DA BD D DA EL BD EL transitive. = and = because corresponding = (CPCTC) thus = by mBDE mLED mLED mELG mBDE mELG, ’s transitive. = because corresponding angles (CPCTC). mB mLEG but then must but it doesn’t… thus the contradiction verifies that must EG EC L ∆DBE ∆LEG be on. e) Proof by Double Translations (Informal Transformational Approach) – This isn’t necessarily proving the relationship but we are to work backwards to show that it works. Translate by vector resulting in ∆ABC ∆BB’D. Translate by vectorresulting in ∆ABC ∆CDE. These two transformations create where is B ∆AB’E, the midpoint of and D is the midpoint of, thus is the midsegment. || because of the translation by vector and = BD because of the translation of to and to AC BD CE. f) by Construction (Informal Investigative Approach) – This isn’t really a proof but it simply a way to check the relationship using a compass and straightedge. Have students construct the midpoints of and , and D E respectively. Create the midsegment. Using compass and straightedge copy onto and BDE DAC
Indirect Proof: That the translation of E by vector will lie on. The way an indirect proof works is to assume the opposite and prove that it creates a contradiction so that the opposite must be true. So let us assume that point L, the image of E translated by vector does not lie on. The contradiction comes when we prove that ∆DBE ∆LEG but then EG must equal EC but it doesn’t…thus the contradiction verifies that L must be on. DA = BD because D is the midpoint and DA = EL because translation vector thus BD = EL by transitive. mBDE = mLED and mLED = mELG because corresponding ’s = (CPCTC) thus mBDE = mELG, by transitive. mB = mLEG because corresponding angles (CPCTC). ∆DBE ∆LEG but then EG must EC but it doesn’t… thus the contradiction verifies that L must be on. e) Proof by Double Translations (Informal – Transformational Approach) This isn’t necessarily proving the relationship but we are to work backwards to show that it works. Translate ∆ABC by vector resulting in ∆BB’D. Translate ∆ABC by vectorresulting in ∆CDE. These two transformations create ∆AB’E, where B is the midpoint of and D is the midpoint of, thus is the midsegment. || because of the translation by vector and BD = because of the translation of AC to BD and to CE. f) by Construction (Informal – Investigative Approach) This isn’t really a proof but it simply a way to check the relationship using a compass and straightedge. Have students construct the midpoints of and , D and E respectively. Create the midsegment. Using compass and straightedge copy BDE onto DAC and