# Clarification of proof

• Aug 15th 2006, 06:56 AM
OReilly
Clarification of proof
Can someone clarify me this proof for this theorem?

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

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

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

I don't understand why we need point \$\displaystyle B_2\$ at all. If we asume that there is point \$\displaystyle C_1\$ such that \$\displaystyle A_1-C_1-B_1\$ ,\$\displaystyle (A,C) \cong (A_1 ,C_1 )\$ and \$\displaystyle (B,C) \cong (B_1 ,C_1 )\$ then it follows \$\displaystyle (A,B) \cong (A_1 ,B_1 )\$ which is the same proof.
• Aug 15th 2006, 11:38 AM
CaptainBlack
Quote:

Originally Posted by OReilly
Can someone clarify me this proof for this theorem?

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

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

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

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

In your second paragraph you have to find a point \$\displaystyle C_1\$ on \$\displaystyle 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 \$\displaystyle B_2\$ you guarantee the existence of both \$\displaystyle C_1\$ and \$\displaystyle B_2\$.

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

RonL