Clarification of proof

Quote:

Originally Posted by OReilly

Can someone clarify me this proof for this theorem?

Theorem: If $A,B,C$ are tri different points of line $l$ and $A_1,B_1,C_1$ are points of line $l_1$ such that $(A,B) \cong (A_1 ,B_1 )$ , then there exists unique point $C_1$ such that $(A,C) \cong (A_1 ,C_1 )$ and $(B,C) \cong (B_1 ,C_1 )$ . Also, point $C_1$ belongs to line $l_1$ and ordering of points $A,B,C$ on line $l$ matches ordering of points $A_1,B_1,C_1$ on line $l_1$ .

I will show proof for order of points $A-C-B$ .

Proof: If $C_1$ and $B_2$ are points of ray $A_1B_1$ such that $A_1-C_1-B_2$ , $(A,C) \cong (A_1 ,C_1 )$ and $(B,C) \cong (B_2 ,C_1 )$ then it follows $(A,B) \cong (A_1 ,B_2 )$ so because of $B_1=B_2$ there is point $C_1$ that meets conditions of theorem.

I don't understand why we need point $B_2$ at all. If we asume that there is point $C_1$ such that $A_1-C_1-B_1$ , $(A,C) \cong (A_1 ,C_1 )$ and $(B,C) \cong (B_1 ,C_1 )$ then it follows $(A,B) \cong (A_1 ,B_1 )$ which is the same proof.

In your second paragraph you have to find a point $C_1$ on $l_1$ that satisfies
two conditions, that is equivalent to finding a real number which satisfies two
equations, there is no apriori guarantee that such a point exists. By
introducing the point $B_2$ you guarantee the existence of both $C_1$ and $B_2$ .

Finally you show that the conditions imply that $B_2=B_1$ .

RonL