Geometry - Help with theorem proof please

[Lee33]

Geometry -- Help with theorem proof please

Homework Statement

Let $A,B,C,D$ be points. If $\vec{AB} = \vec{CD}$ then $A=C$.

Homework Equations

None

The Attempt at a Solution

This question was a theorem in my book that wasn't proved. I am wondering how to prove it?

It is saying that the vertex $A$ must equal $C$ if the ray $\vec{AB} = \vec{CD}$.

The definition I have for ray is:

$\vec{AB} = \vec{AB} \cup \{ C \in P \ | \ A-B-C\}.$ Where $A-B-C$ means $B$ is between $A$ and $C$. And $P$ is the set of points.

 

[tiny-tim]

Hi Lee33!

(your definition doesn't look quite correct)

Suppose A ≠ C

A is in $\vec{CD}$, so … ? ​

 

[Lee33]

tiny-tim - Can you elaborate a bit more please?

 

[tiny-tim]

Hi Lee33!

Apply the definition you were given …

The definition I have for ray is:

$\vec{AB} = \vec{AB} \cup \{ C \in P \ | \ A-B-C\}.$ Where $A-B-C$ means $B$ is between $A$ and $C$. And $P$ is the set of points.

Suppose A ≠ C

C is in $\vec{AB}$, so what can you say about A B and C ? ​

 

[Lee33]

If C is in $\vec{AB}$ and $C\ne A$ then B is between A and C?

 

[Mark44]

The definition I have for ray is:

$\vec{AB} = \vec{AB} \cup \{ C \in P \ | \ A-B-C\}.$ Where $A-B-C$ means $B$ is between $A$ and $C$. And $P$ is the set of points.

This definition makes no sense to me. First off, why would $\vec{AB}$ be equal to itself union some other thing (unless the other thing happened to be the empty set).

Second, how do you interpret $\{ C \in P \ | \ A-B-C\}$? Does | have its usual meaning of "such that" or am I missing something? An explanation, in words, would be helpful.

Third, where are these points? Are they on a line or are they in the plane?

Fourth, how do you get that A - B - C means that B is between A and C?

 

[Lee33]

Sorry, I will elaborate.

First question: If A and B are distinct points in a metric geometry then the line segment from A to B is the set $\vec{AB}=\{C \in P \ | \ A-C-B \ or \ C = A \ or \ C = B\}$.

If A and B are distinct points in a metric geometry then the ray from A toward B is the set $\vec{AB}=\vec{AB}\cup \{C\in P \ | \ A-B-C\}.$

Second question: Yes, it means such that. Let P be the set of points in a metric geometry, and let C be a point in P such that B is between A and C.

Third question: They are on a line.

Fourth: That is just a notation for convenience. $A-B-C$ just means B is between A and C.

I will add the definition of between-ness: B is between A and C if the distance $d(A,B)+d(B,C) = d(A,C)$.

 

[tiny-tim]

Hi Lee33!

(just got up :zzz:)

If C is in $\vec{AB}$ and $C\ne A$ then B is between A and C?

nooo, C is (strictly) between A and B

ok, and if A is in $\vec{CD}$, then … ? ​

 

[Lee33]

If A is in $\vec{CD}$ then A is between C and D.

 

[tiny-tim]

If A is in $\vec{CD}$ then A is between C and D.

yes (strictly between)

ok, now you have two statements, and you should be able to prove a contradiction (thereby showing that "A ≠ C" was false)

(drawing yourself a diagram might help)

 

[Lee33]

Alright thanks for the help! I will use your hints.

Question. Do I use both statements in my proof? That is, suppose $A\ne C$ and A is in $\vec{CD}$ then A is bewteen C and D. Also, I will use if $A\ne C$ and C is in $\vec{AB}$ then C is between A and B?

 

[tiny-tim]

Question. Do I use both statements in my proof? That is, suppose $A\ne C$ and A is in $\vec{CD}$ then A is bewteen C and D. Also, I will use if $A\ne C$ and C is in $\vec{AB}$ then C is between A and B?

yes

 

[gopher_p]

You are using the same notation for line segment and ray, and it's confusing the bejeesus out of the people who are trying to help you.

Might I suggest $\overline{AB}$ for the segment and $\overrightarrow{AB}$ for the ray so that $\overrightarrow{AB}=\overline{AB}\cup \{C\in P \ | \ A-B-C\}.$

 

[tiny-tim]

… it's confusing the bejeesus out of the people who are trying to help you.

it's not confusing me

 

[gopher_p]

it's not confusing me

Are you sure? Like, really sure? Because when Lee asked

If C is in $\vec{AB}$ and $C\ne A$ then B is between A and C?

you replied

nooo, C is (strictly) between A and B

which is generally false regardless of which of Lee's two definitions of $\vec{AB}$ you're using. :tongue:

 

[tiny-tim]

… which is generally false …

well, Lee33 didn't contradict me, sooo i assume i got it right!

 

[Lee33]

gopher_p - Sorry about that, you're right!

tiny-tim - If $A\ne C$ then $C\in \vec{AB}$ thus $A-C-B$ but where will the point $D$ be?

 

[tiny-tim]

but you haven't used …

If A is in $\vec{CD}$ then A is between C and D.

 

[Lee33]

So my proof should go like this:

Suppose $A\ne C$, now since $\vec{AB}=\vec{CD}$ then $A\in \vec{CD}$ and $C\in \vec{AB}$. Thus $C-A-D$ and $A-C-B$ which is a contradiction?

 

[tiny-tim]

yes!

if I'm understanding the terminology correctly, you can't have both $A-C$ and $C-A$ unless C = A

 

[Lee33]

Thank you very much for the help!