3. Justify each step in the following proof of Proposition I(SEGMENT SUBTRACTION). IFA' B *C.D'E'F. AB= DE, and ACDF, then BCEFI PROOF: (1) Assume on the contrary that BC is not congruent to EF. (2) Then there is a point G on Ef such that BC e EG. (3) G+ F. (4) Since AB DE, adding gives AC DG. (5) However, AC DF. (6) Hence, DF DG. (7) Therefore, F G. (8) Our assumption has led to a contradiction; hence, BC EF.

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter1: Line And Angle Relationships
Section1.4: Relationships: Perpendicular Lines
Problem 12E: In Exercise 11 and 12, provide the missing statements and reasons. Given: 12;34 s 2 and 3 are...
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Please answer number 3

1. Given A'B'Cand A C'D.
a. Prove that A, B, C. and Dare four distinct points.
b. Prove that A, B. C. and Dare collinear.
2. Given A'B'C. If P is a fourth point collinear with A, B, and C, use Proposition (Given A' B
c and A'c D. Then B'c Dand A' B D) and an axiom to prove that ~A B P= -A•
C• P.
3. Justify each step in the following proof of Proposition I(SEGMENT SUBTRACTION). IF A' B
*C.D'E'F, AB= DE, and AC DF, then BCEFI
PROOF:
(1) Assume on the contrary that BC is not congruent to EF.
(2) Then there is a point G on Ef such that BC e EG.
(3) G+ F.
(4) Since AB DE, adding gives AC DG.
(5) However, AC DF.
(6) Hence, DF DG.
(7) Therefore, F = G.
(8) Our assumption has led to a contradiction; hence, BC EF.
Transcribed Image Text:1. Given A'B'Cand A C'D. a. Prove that A, B, C. and Dare four distinct points. b. Prove that A, B. C. and Dare collinear. 2. Given A'B'C. If P is a fourth point collinear with A, B, and C, use Proposition (Given A' B c and A'c D. Then B'c Dand A' B D) and an axiom to prove that ~A B P= -A• C• P. 3. Justify each step in the following proof of Proposition I(SEGMENT SUBTRACTION). IF A' B *C.D'E'F, AB= DE, and AC DF, then BCEFI PROOF: (1) Assume on the contrary that BC is not congruent to EF. (2) Then there is a point G on Ef such that BC e EG. (3) G+ F. (4) Since AB DE, adding gives AC DG. (5) However, AC DF. (6) Hence, DF DG. (7) Therefore, F = G. (8) Our assumption has led to a contradiction; hence, BC EF.
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