I have to prove that if A-B-C and A-D-C, then A,B,C,D are points of one line and it isn't B-A-D.
I see that B and D are on same side based on A-B-C and A-D-C, so it couldn't be B-A-D.
Solution from book goes: If we assume that B-A-D is true then B and D are on different sides of A, so based on A-B-C it will be A-C-D which is false.
I don't understand why solution from book says that based on B-A-D and A-B-C it must be A-C-D?
Isn't that based on B-A-D and A-B-C must be D-A-C?
I can explain some of the logic in the text.
In assuming B-A-D and given A-B-C it follows: from B-A-D that B & D are on opposite sides of A; from A-B-C it follows that B & C are on the same side of A. Therefore, it follows that C & D are on opposite sides of A which contradicts the given A-D-C.
Again there are many, many different sets of ‘betweeness axioms’. So the above may be at odds with your text. However it should be close.
By the way, there is a good and different proof in Ed Moise’s geometry book in chapter 8.
See, the problem with Euclidean Geometry is that it is not fully formal (though close to it).
A Late 19th -20th Century mathemation David Hilbert worked on a problem of making Euclidean geometry more formal. It became know as Hilbert's Axioms I belive he used the 5 original postulates (with the revision of the Playfair Axiom) and 16 of his own (actaully the last one can be proven from the other 15 thus it got later omitted because it can lead to a risk in creating a non-consistent system). Hilbert concentrates greatly on point order. Meaning the name of point by accepting "betweenness" as an undefinable term.
This is why I love math so much. Any non-mathemation will look at these axioms of Hilbert and think he is retarded and what purpose is to study it (there is none!). Ah! Pure mathematics can it never have a purpose, Amen (Hardy).