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 February 10th, 2014, 10:29 PM #1 Newbie   Joined: Jan 2014 Posts: 16 Thanks: 0 Line segment If $\vec{AB}=\vec{CD}$ (line segments) in a metric geometry, prove that $A=C$ and $B=D.$ How can I prove this?
February 11th, 2014, 12:41 PM   #2
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Re: Line segment

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 Originally Posted by tom33 If $\vec{AB}=\vec{CD}$ (line segments) in a metric geometry, prove that $A=C$ and $B=D.$ How can I prove this?
You need to clarify your notation. Is it possible that A=D and B=C?

 February 11th, 2014, 01:45 PM #3 Newbie   Joined: Jan 2014 Posts: 16 Thanks: 0 Re: Line segment Well, the book didn't elaborate on that. I assume since it didn't elaborate then A=D and B=C is not applicable.
 February 12th, 2014, 03:51 PM #4 Global Moderator   Joined: May 2007 Posts: 6,581 Thanks: 610 Re: Line segment How is line segment defined in the book?
February 13th, 2014, 04:45 AM   #5
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Re: Line segment

Quote:
 Originally Posted by tom33 Well, the book didn't elaborate on that. I assume since it didn't elaborate then A=D and B=C is not applicable.
Whether it was "elaborated" or not, to do a problem like this you will have to know the definition of a line segment and what it means for two line segments to be equal.

 February 13th, 2014, 05:26 PM #6 Newbie   Joined: Jan 2014 Posts: 16 Thanks: 0 Re: Line segment The definition I have, in the book, for line segment is: If A and B are distinct points in a metric geometry then the line segment from A to B is the set $\vec{AB}=\{ C\in P \ | \ A-C-B \ or \ C = A \ or \ C = B\}.$ Where P is the set of points and A-C-B means that the point C is between A and B.
 February 15th, 2014, 12:19 AM #7 Newbie   Joined: Jan 2014 Posts: 16 Thanks: 0 Re: Line segment Do you have any idea on how I can prove this?
 February 15th, 2014, 12:44 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,503 Thanks: 1739 What exactly does $\small A-C-B$ mean, i.e., what does "between A and B" mean?
 February 16th, 2014, 02:46 PM #9 Newbie   Joined: Jan 2014 Posts: 16 Thanks: 0 Re: Line segment C is between A and B if the distance $d(A,C)+d(C,B)= d(A,C).$

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