In Plane and solid geometry by Fletcher Durell(p.32), it states that:

A triangle is a portion of plane bounded by three straight lines, as the triangle ABC".

How does this statement relates with the existence of a triangle. For example, how could three sides be chosen so the figure obtained is a portion of a plane bounded by three straight lines?

  • 2
    $\begingroup$ I don't know what your question is. That definition tells you what the author means by the word "triangle". Presumably the "$ABC$" in your quote refers to an example (a picture). $\endgroup$ – Ethan Bolker Jul 16 '18 at 22:23
  • $\begingroup$ You just have to choose three lines, no two of which are parallel. Is this what you're asking? $\endgroup$ – saulspatz Jul 16 '18 at 22:58
  • $\begingroup$ @EthanBolker: In simple words, having three lines, how a triangle be made where the lines could take any measure? $\endgroup$ – justin Jul 17 '18 at 21:35
  • $\begingroup$ @EthanBolker: I'm sorry. I'm rephrasing the comment. In simple words, how could a triangle be made using three lines of any measure? $\endgroup$ – justin Jul 18 '18 at 21:05
  • $\begingroup$ You keep trying, but I still don't understand the question. If you have three line segments and the sum of two of the lengths is less than the length of the third then they can make a triangle. If this isn't what you mean there's nothing more I can do to help. $\endgroup$ – Ethan Bolker Jul 18 '18 at 21:33

In order for a portion of the plane to be bounded by three straight lines, the lines in question need only:

  • not be parallel to each other
  • intersect in more than one point (i.e. 3 points; if you wish the area of the bounded region to be nonzero)*

*Proofs available upon request.


As a construction: How to make a triangle (in a plane):

Choose (or draw) 2 non-parallel lines. These lines intersect in a point.

Choose an additional line that:

  • Does not intersect the aforementioned point. (unless you want the bounded area to be zero)
  • Is not parallel to either of the prior 2 lines.

For example, in the Cartesian plane: If you use the x and y axis for the first two lines, then they intersect in the origin. If the third line you choose isn't parallel to either axis, and doesn't run through the origin, it can be written in the form y=ax+b or x=cy+d, such that a, b, c, and d are $\neq 0$.

For another construction:

Choose two non-parallel lines. Pick a point on each line, other than the intersection, and draw a line through those points. The area bounded by the lines ought to be between the third line you draw, and the intersection (of the first two lines you chose).

There are some illustrated examples here: https://en.wikipedia.org/wiki/Parallel_postulate and here: http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf under Proposition 1. I understand this book (Euclid's Elements) is filled with constructions and constructive proofs.

A Pictures of 3 lines, no 2 of which are parallel: a triangle Note that if all three lines intersected in the same place, there wouldn't be any bounded region of non-zero area.

  • $\begingroup$ Yeah please give me proof for the (area of bounded region - nonzero). I hope this is what makes a three line figure to exist. $\endgroup$ – justin Jul 17 '18 at 21:44
  • $\begingroup$ For a proof, I need starting assumptions. Is it clear that in order to be a region of zero area, the region would have to be a point or a line? $\endgroup$ – user558317 Jul 24 '18 at 0:05
  • $\begingroup$ Yes of course. I hope an area should be bounded by at least 3 lines. $\endgroup$ – justin Jul 24 '18 at 21:17
  • $\begingroup$ At least three? The question as stated, concerns only the necessary conditions for triangles, not the minimum necessary number of lines to make a region of non-zero area (which is also 3). $\endgroup$ – user558317 Jul 26 '18 at 19:21
  • $\begingroup$ More precisely, how to choose three lines(side length) so it intersects. I hope this should be analyzed by checking whether the two sides intersect at a point by knowing their loci. $\endgroup$ – justin Jul 26 '18 at 21:19

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