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I am currently using the book "Euclidean Geometry" by David M. Clark.

I have to prove that: two lines parallel to the same line are parallel to each other.

I am not allowed to use angle measure yet (degrees).

I am able to use any triangle congruence (SSS, SAS, AAS, ASA, HL)

I am allowed to use angle bisectors, midpoints, circles, right angles, isosceles triangles, vertical angles, corresponding angles, alternate interior angles, exterior angles, and squares to prove this.

We have these theorems which may be useful in proving this:

  • If two lines have a transversal which forms alternative interior angles that are congruent, then the two lines are parallel.

  • If two lines have a transversal which forms corresponding angles that are congruent, then the two lines are parallel.

  • An exterior angle of a transversal is not congruent to either
    opposite interior angle.

  • Vertical angles are congruent.

  • Congruent angles have congruent supplements.

  • All right angles are congruent.

  • Axiom 5: For every line l and every point P not on l, there is at most one line containing P that is parallel to l.

I started off drawing two parallel lines, l and m and point P on m, but I really don't even know where to begin.

Any help would be appreciated.

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    $\begingroup$ If you suppose the two lines are not parallel and so are incident, then you have a contradiction with Axiom 5. $\endgroup$ – Tony Piccolo Oct 15 '15 at 21:25
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    $\begingroup$ @TonyPiccolo Would the contradiction then be that the two lines both intersect at point P because then two lines would be containing P, which contradicts what Axiom 5 says? $\endgroup$ – Mathgirl Oct 16 '15 at 1:36
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    $\begingroup$ You would have two distinct lines such that $\dots$ Axiom says at most one ! $\endgroup$ – Tony Piccolo Oct 16 '15 at 5:12
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I think I have a correct proof now.

Proof by contradiction: Assume to the contrary that two lines parallel to the same line are not parallel to each other. Without loss of generality, assume line m and line n are parallel to a line l, but m and n are not parallel to each other. Then, m and n intersect at a point, P that is not on line l. However, this contradicts Axiom 5 because two lines would be containing P and be parallel to l. So the assumption that m and n are not parallel was incorrect. Thus, m and n are parallel to l and also parallel to each other.

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Draw a line parallel to A as B . Alternate angles a = b

Draw a line parallel to A as C . Alternate angles a = c

a = b = c

$ \because$ a = c, A is parallel to C by the converse thm, the first one in the given list.

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  • $\begingroup$ By the converse of what Theorem? $\endgroup$ – Mathgirl Oct 15 '15 at 20:59
  • $\begingroup$ as given in edit above $\endgroup$ – Narasimham Oct 16 '15 at 20:54

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