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On the fourth page of Simmons' Precalculus Mathematics In a Nutshell some basic postulations for triangles are provided such as: corresponding angles of parallel lines are equal as well as their alternate interior angles, the sum of the angles of a triangle equals 180°, etc. But one proposition he made was that an exterior angle was equal to the sum of the opposite interior angles (remote angles?) which he very nonchalantly says in passing. I attempted proving it which can be seen in the provided picture on the top which I explain in words on the right side saying "Exterior angle C is associated with two angles, therefore its sum is equal to 60° + 60° = 120°, the sum of the opposite interior angles." And another proof on the bottom given by Oria Gruber, a user on this site whose answer I was unsatisfied with. [ The original question for that answer Are exterior angles equal to the sum of two remote angles? Please help explain. ] [My work (on top) and reiteration of another user's work [Oria Gruber]1 (on bottom)]2

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  • $\begingroup$ Which proposition came first? A proof is only valid if the things it depends on are already known (definitions or proved propositions). And you did not say why the proof you linked to was unsatisfactory. $\endgroup$ – David K Dec 20 '18 at 16:18
  • $\begingroup$ @David K These come with figures but I doubt I can post a picture of them straight from the book: 1) "One degree is one-nineteenth of a right angle", 2) "If a transversal is drawn across a pair of parallel lines then corresponding angles are equal and alternate interior angles are equal." 3) "The sum of the angles in any triangle equals 180." 4) "As a direct consequence... the sum of the acute angles in a right triangle equal 90" 5) "in any triangle an exterior angle equals the sum of the opposite interior angles." $\endgroup$ – Voisistere Dec 20 '18 at 16:51
  • $\begingroup$ The question is already a bit hard to follow and it does not help to have to refer back and forth between the question and the comment. You could try to edit the question to make it easier to see what you are writing about. $\endgroup$ – David K Dec 20 '18 at 18:20
  • $\begingroup$ Regarding the two proofs, this is supposed to be a general theorem about all triangles, not just when they have 60-degree angles; and I think in your notes you have misrepresented the other proof somewhat by putting the labels "theorem" and "proof" arbitrarily on two steps of the proof and omitting the last line of the proof completely. There is a small bit of algebra you are expected to do to get from the first two lines to the last one in that proof; is that what you're missing? $\endgroup$ – David K Dec 20 '18 at 18:23
  • $\begingroup$ "and I think in your notes you have misrepresented the other proof somewhat by putting the labels "theorem" and "proof" arbitrarily on two steps of the proof and omitting the last line of the proof completely." A reasonable criticism of my notes, but not much was said in the proof which was my main issue with it. It also wasn't very thorough, stating first that Angle 1 + Angle 2 + Angle 3 = 180 then jumping to Angle 3 + Angle 4 = 180 which does not intuitively follow from his first statement. Why is Angle 3 + Angle 4 = 180 true? Regardless, I now know the answer provided by user3482749. $\endgroup$ – Voisistere Dec 20 '18 at 20:54
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In Elements I, 32, Euclid proves the angles of a triangle sum to "two right angles" ($180^o$) by first showing that an exterior angle equals the sum of the two opposite interior angles. So he can't use the former to prove the latter.

Instead, he shows that exterior $\angle ACD=\angle ABC+\angle BAC$ by drawing$$CE\parallel BA$$and observing that, by I, 29, alternate interior$$\angle BAC=\angle ACE$$and corresponding$$\angle ABC=\angle ECD$$Therefore, the whole exterior $\angle ACD=\angle BAC+\angle ABC$.

Only then does Euclid prove--almost as an afterthought--the theorem fundamental in his geometry, that the angles of a triangle sum to $180^o$, or "two right angles" as he puts it. exterior angle of triangle

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  • $\begingroup$ This is the only reasonable answer - the only one that bothers to consider which results we're taking as axioms and which must be proven. (Although in "Simmons' Precalculus Mathematics In a Nutshell" other axioms may be the starting ones.) $\endgroup$ – Misha Lavrov Dec 31 '18 at 16:37
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Call the interior angles $A$, $B$, and $C$. Then the exterior angles are $180 - A$, $180 - B$, and $180 - C$. We know that $A + B + C = 180$, so $A = 180 - B - C$, $B = 180 - A - C$, and $C = 180 - A - B$. Substutituing these into our expressions for the exterior angles, our exterior angles at $A$, $B$, and $C$ respectively are $180 - (180 - B - C) = B + C$, $180 - (180 - A - C) = A + C$, and $180 - (180 - A - B) = A + B$.

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By angle sum property, $A+B=180°-C$ (which is nothing but the exterior angle at $C$)

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In a ΔABC the sum of all the angles is 180°.

∠A+∠B+∠C=180°

Hence, ∠A+∠B = 180°-∠C

BC is extended to D. ∠ACD+∠C=180° [Linear pair]

∠ACD=180°-∠C

∠A+∠B=∠ACD=180°-∠C

Hence, the sum of opposite interior angles is equal to the exterior angle.

Proved

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