I'm trying to essentially prove that the interior angles of a triangle add up to 180 degrees. However, I'm trying to prove it without mentioning measurements of an angle. I think I understand the proof, but I'm having a lot of trouble with the syntax and how to word even what I'm trying to prove. Will somebody give me advice on how to improve my proof? The definition of the word "supplementary" that I'm using says that two angles are supplementary if they share on side and the other sides are opposite rays. Here it is:

1) Given a triangle ABC (by incidence Axioms 3 and 1) we arbitrarily choose vertex A, and (by Euclidean's Parallel Postulate) find a line through A parallel to BC. Let l be the line through A such that $l \parallel BC$.

2) Let D be a point such that C*A*D (by Betweenness Axiom 2) and E be a point such that B*A*E (by Betweenness Axiom 2).

3) Let ray AF be a ray emanating from A such that F is on line l, and F and D are on the same side of line EA (by Betweenness Axiom 4) and let ray AG be the opposite ray to ray AF such that G and E are on the same side of line DA. (Betweenness Axiom 4)

4) Then DC and BE are lines transversal to l and $\line BC$. (by definition)

5) $\angle{ABC} \cong \angle{BAF}$ and $\angle{ACB} \cong \angle{CAG}$ (alternate interior angles).

6) Also, $\angle{BAC} \cong \angle{BAC}$ (by reflexivity of Congruence Axiom 2).

7) Then $\angle{DAG}$ is supplementary to $\angle{FAG}$, and $\angle{FAE}$ is supplementary to $\angle{EAG}$. (by definition)

8) ray AD is interior to $\angle{FAE}$ and ray AE is interior to $\angle{DAG}$ (by step 3)

9) by angle addition, $\angle{FAG} \cong \angle{FAG}$.

10) F*A*G by construction, so by Betweenness Axiom 1, GAF is a straight line.

Therefore a straight line is congruent to the sum of the interior angles of a given triangle.

I know, it needs work. How do I word what I'm trying to prove? I don't want to say that the sum of the interior angles is a straight line because they are not all on the same line. Does the way I worded it make sense?

You may want to draw it out to understand it. I don't know how to attach a diagram. Thanks!

[I'm working on editing the question. Need to go find some mathjax codes...]


enter image description here

Look at the 3 angles. They form a straight line at point B and are equal on the triangle because of the parallel lines.

I lack the English terms, but I hope it's clear enough

  • $\begingroup$ Thank you for providing a diagram for my proof. I'm trying to prove that this is true without mentioning the 180 degrees. I'm mostly having trouble with the wording of it. $\endgroup$ – Amber Oct 17 '13 at 21:10
  • $\begingroup$ I passed through B a line that is parallel to AC. Then the two red angles are on the same sides of a line crossing two parallel lines (name please) therefore they are equal. The exact same applies to the two orange angles. The two purple angles are equal because they are the opposite sides of the same angle (name). The three angles form a straight line and they are the equal to the angles in a triangle. This is without mentioning degrees. $\endgroup$ – DannyDan Oct 17 '13 at 21:13

Diagram from wikibooks may be useful just after Example 1: http://en.wikibooks.org/wiki/High_School_Trigonometry/Angles_in_Triangles#Triangles_and_Their_Interior_Angles
and attaching it in your post could be like this: straight angle


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.