Proving lines are parallel

[MathMathter]

If I'm given that AB is parallel to FI, how would I go about proving that GR is parallel to FI in a formal two-column proof? It appears that GR and AB are the same line, but I don't know if we can just assume that. And if we can assume it, I don't know what the formal statements and reasons would be to get to GR || FI. Thanks!

 

[pka]

View attachment 3645
If I'm given that AB is parallel to FI, how would I go about proving that GR is parallel to FI in a formal two-column proof? I know GR and AB are the same line, but I don't know how to formally state that. Thanks!

I am sure it is because I taught axiomatic geometry for over thirty years, I find it hard to think that any competent text and/or instructor would ask this question. Much less ask for the awful two-column proof.

Two points determine a line. You are given $\displaystyle G - A - B - H$ (betweenness).
Therefore, $\displaystyle \overleftrightarrow {AB} = \overleftrightarrow {GH}$.

You are given $\displaystyle \overleftrightarrow {AB}\parallel \overleftrightarrow {FI}$ so $\displaystyle \overleftrightarrow {GH}\parallel \overleftrightarrow {FI}$.

 

[MathMathter]

The context, here, is that ABIF is a parallelogram (imagine segments AF and BI in the diagram), and I need to prove that GRIF (imagine segments GF and RI in the diagram too) is also a parallelogram. I'm given that GF || RI and I know that AB || FI since ABIF is a parallelogram. I want to show that GR || FI too so that I can prove that GRIF is a parallelogram.

I appreciate the response, pka, but it doesn't quite answer my questions (and your letters are off).

 

[pka]

The context, here, is that ABIF is a parallelogram (imagine segments AF and BI in the diagram), and I need to prove that GRIF (imagine segments GF and RI in the diagram too) is also a parallelogram. I'm given that GF || RI and I know that AB || FI since ABIF is a parallelogram. I want to show that GR || FI too so that I can prove that GRIF is a parallelogram.

You failed to post a complete question. You did not give us all of the facts that were given to you. This should be a lesson to you. In the further post the entire problem. We are not mind readers here.

Now as it turns out, I gave you the answer to the actual question.
Although we do it all the time, strictly speaking only lines are parallel and not line segments.
If we have two parallel lines then two line segments, one on each line we can loosely say that the segments are 'parallel'. If they have the same length then they form a parallelogram.

 

[MathMathter]

Whoa...no need for the mind-reading lecture. The problem was complete enough as it was originally posted. I just provided the context in response to your remark about the competence of the instructor.

So this is what I'm seeing in terms of a two-column proof...statement: [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]←[/FONT][FONT=MathJax_Main]→[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]G[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main]←[/FONT][FONT=MathJax_Main]→ (looks like copy/paste didn't work here...the arrows should be above the letters to indicate that they're lines)[/FONT], reason:betweenness. But I thought betweenness (of points) was for something like "If A-B-C, then AB + BC = AC". Then for the statement [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]←[/FONT][FONT=MathJax_Main]→[/FONT][FONT=MathJax_Main]∥[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Math]I[/FONT][FONT=MathJax_Main]←[/FONT][FONT=MathJax_Main]→[/FONT] so [FONT=MathJax_Math]G[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main]←[/FONT][FONT=MathJax_Main]→[/FONT][FONT=MathJax_Main]∥[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Math]I[/FONT][FONT=MathJax_Main]←[/FONT][FONT=MathJax_Main]→[/FONT], there was no reason provided. I'd guess "substitution", but I've never seen it in this context.