Prove that the sum of any two angles of a triangle is less than $180$ degrees without the notion of a parallel line.

Question

This question is really puzzling me and all the "proofs" which I did were intuitively using the notion of a parallel line somewhere.
I am starting to think that this is impossible but I just want to be absolutely sure.
Please Help!

Answer 1

Note: This proof is from Theorem 1, Corollary 2 in the lecture notes mentioned in Reference 1.

Transcription of key arguments below:

The angle sum of a triangle depends on the choice of Parallel Postulate and, given such a choice, is always less than, equal to, or greater than $180◦$.

On the other hand, if we only assume Euclid’s first four postulates, along with the axioms of incidence, congruence, continuity and betweenness, the angle sum of a triangle is always less than or equal to 180◦. This geometry is called neutral (or absolute) geometry.

Theorem 1 (Exterior Angle Inequality) The measure of an exterior angle of a triangle is greater than the measure of either remote interior angle

Proof. Given $\triangle ABC$, extend side $BC$ to ray $\overrightarrow {BC}$ and choose a point $D$ on this ray so that $B − C − D$.

Claim: $m∠ACD > m∠A$ and $m∠ACD > m∠B$. Let $M$ be the midpoint of $AC$ and extend the median $BM$ so that $M$ is the midpoint of $BE$.

Then $∠AMB$ and $∠CME$ are congruent vertical angles and $\triangle AMB \cong \triangle CME$ by SAS. Consequently, $m∠ACE = m∠CAB$ by CPCTC. Now, $E$ lies in the half-plane of $A$ and $\overleftrightarrow {CD}$ since $A$ and $E$ are on the same side of $\overleftrightarrow {CD}$. Also, $E$ lies in the half-plane of $D$ and $\overleftrightarrow {AC}$ since $D$ and $E$ are on the same side of $\overleftrightarrow {AC}$. Therefore $E$ lies in the interior of $∠ACD$, which is the intersection of these two half-planes. Finally, $m∠ACD = m∠ACE + m∠ECD > m∠ACE = m∠CAB = m∠A$. The proof that $m∠ACD > m∠B$ is similar.

Corollary 2 The sum of the measures of any two interior angles of a triangle is less than $180◦$.

Proof. Given $\triangle ABC$, extend side $B$C to $\overleftrightarrow {BC}$ and choose points $E$ and $D$ on $\overleftrightarrow {BC}$ so that $E −B −C − D$ as given in figure below:

By Theorem 1, $m∠A<m∠ACD, m∠B<m∠ACD$, and $m∠A<m∠ABE$. By adding $m∠C = m∠ACB$ to both sides of the first two inequalities, and by adding $m∠B = m∠ABC$ to both sides of the third we obtain $$m∠A + m∠C<m∠ACD + m∠ACB = 180◦$$

$$m∠B + m∠C<m∠ACD + m∠ACB = 180◦$$

$$m∠A + m∠B<m∠ABE + m∠ABC = 180◦$$

Hence the proof.

References:

1: *, Lecture Notes from MATH 353-Survey of Geometry, Millersville University. Fall 2007.

2: Marvin Greenberg, “Euclidean and Non-Euclidean Geometries: Development and History,” 3rd Ed., W.H. Freeman and Co., New York, 1993.

Answer 2

I didn't understand the meaning of "Notion of parallel lines..." though I had a thought:

Since the sum of the angles, (say $\angle A,\angle B, \angle C$) of a triangle is $180^\circ$ so the sum of any two angles will be $180^\circ - $the third angle $<180^\circ$. Isn't that sufficient? Or did I missed something?

Answer 3

Hint: Each of the sides is a transversal of the two others. On which side of this transversing will the two other sides meet?

Answer 4

Maybe you need to use the property that sum of two interior angles of a triangle is equal to the opposite exterior angle.

Consider any $\Delta ABC$. Then, $$\angle A+\angle B=\angle ACD<180^o,$$ since, $BD$ is a straight line and $\angle ACD<\angle BCD=180^o$.

Answer 5

If we'll assume that for any triangle a sum of three measured angles is equal to $x$, we can get the following.

Let $D\in AC$ for $\Delta ABC$.

Thus, $$2x=(\measuredangle ABD+\measuredangle BDA+\measuredangle BAD)+(\measuredangle DBC+\measuredangle BCD+\measuredangle BDC)=$$ $$=x+\measuredangle BDA+\measuredangle BDC=x+180^{\circ},$$ which gives $$x=180^{\circ}.$$ But we can not assume it. We need an axioma of parallels for the right proof.