36

I'm looking for a way to look at a triangle, and perhaps visualize a few extra lines, and be able to see that the interior angles sum to $180^\circ$.

I can visualize that supplementary angles sum to $180^\circ$. I'd like to be able to see the interior angle sum similarly...

I can see that the exterior angles must sum to $360^\circ$, because if you walked around the perimeter, you would turn around exactly once (though I can tell this is true, I don't really see it). I also saw a proof on KA, where the exterior angles were superimposed, to show they summed to $360^{\circ}$ (though I'm not 100% comfortable with this one).

Finally, for $a$, $b$, and $c$ exterior angles $a+b+c=360$:

\begin{align} (180-a) + (180-b) + (180-c) & = 3\times 180 - (a+b+c) \\ & = 3\times 180 - 360 \\ & = 180 \\ \end{align}

But I find this algebra hard to see visually/geometrically. Is there proof that enables one to directly see that the interior angles of triangle sum to $180^\circ$?

A couple of secondary questions:

  • am I visually deficient in my ability to imagine?
  • or, am I asking too much of a proof, that I be able to see it, and that beimg able to tell that it is true should be enough...?
hyperpallium
  • 953
  • 7
  • 16
  • 1
    The typical proof introduces a line through a vertex, parallel to the opposite side. – Blue Apr 08 '18 at 12:39
  • 1
    Draw a parallel line (how?) and consider internal alternate angles. – Piquito Apr 08 '18 at 12:42
  • 2
    There is a very nive way of animating this, a line starts along one side, moves to a vertex, rotates through the interior angle until it lies along the next side, moves along that side to the next vertex, rotates to lie along the third side, moves to the last vertex, and rotates back into its original position, only rotated by 180°. I took a quick look on Youtube, but couldn't find it, I will keep looking – Malcolm Apr 08 '18 at 12:51
  • This is basically the same as the answer from @Nij. Visualize yourself walking along the base of the triangle. At the first vertex rotate by the amount of the internal angle. Now continue to the next vertex: note that you are walking backwards! At the next vertex, again rotate by the amount of the internal angle. Continue to the next vertex. You will not be walking forwards. At the third vertex, rotate by the internal angle and head to the first vertex. You are now traversing the base backwards. You have been rotated by 180 degrees. Yay! – Hugh Meyers Apr 09 '18 at 13:39
  • This is a common activity for school kids to help illustrate this idea. I just did a quick search on youtube and found this [one.](https://www.youtube.com/watch?v=vw-rOqDBAvs) This is the same as many of the answers, just done with a piece of paper. – N. Owad Apr 12 '18 at 01:28
  • If one break a triangle, triangle is equivalent to just line. So it's 180 degree. – Takahiro Waki Apr 12 '18 at 19:06

13 Answers13

130

Since that fact about the angle sum is equivalent to the parallel postulate, any visualization is likely to include a pair of parallel lines. Here's one from wikipedia:

enter image description here

Ethan Bolker
  • 80,490
  • 6
  • 95
  • 173
63

Ethan Bolker's answer is the standard proof, but a more visual way to see the result is to tile the plane with copies of your triangle. It is so effective, I will not even include a drawing.

Just think about it... the tiling is made by three sets of parallel lines in the directions of the triangle sides, and they meet at vertices of the tiling, where you will then find two copies of each angle.

Edit. By popular request, there's a picture below:

Edit 2. I like Steven Gubkin's answer better :) That is what I used to call the "near-sighted method"

enter image description here

Rodrigo A. Pérez
  • 1,353
  • 1
  • 8
  • 13
  • 6
    I think this answer is better than mine. – Ethan Bolker Apr 08 '18 at 15:21
  • 10
    Hate to be that guy, but I can't see it. Could someone include an image? – user1717828 Apr 09 '18 at 01:45
  • 1
    @user1717828 Look at [this](https://upload.wikimedia.org/wikipedia/commons/thumb/2/23/Triangle_Tiling.svg/700px-Triangle_Tiling.svg.png). Draw a small circle in a point, the sum of all the angles must be $360$. There are twice each angle of the original triangle, then you must have $360=2a+2b+2c$ and hence $a+b+c=180$. – Red Banana Apr 09 '18 at 02:29
  • 2
    @user1717828 I added an image. The poster may remove it if he wants to leave it to your imagination. – Ethan Bolker Apr 09 '18 at 14:25
  • This is essentially the same as Ethan Bolker's answer but with the picture reflected and tiled. Does it really add anything?... – Jam Apr 09 '18 at 18:17
  • 2
    @Jam: *Theoretically*, they're equivalent (as they must be), so no. But at issue here is whether they have different pedagogical values. This one might *seem* more direct to some readers. Certainly Ethan himself seems to find that plausible. – Brian Tung Apr 09 '18 at 21:56
26

This is very similar to Ethan Bolker's and Rodrigo A. Pérez's answers, but I made a small animation to illustrate a version that I like.

enter image description here

jkej
  • 377
  • 2
  • 6
  • 2
    Interesting. I don't have the time to track it right now, but I'm curious how this proof is making use of the parallel postulate. – Paul Sinclair Apr 09 '18 at 12:32
  • very beautiful and understandable. – Mohammad Riazi-Kermani Apr 09 '18 at 13:32
  • 5
    @PaulSinclair SAS similarity is equivalent to the parallel postulate. See Petrunin's notes on geometry for example. – Steven Gubkin Apr 09 '18 at 13:52
  • 1
    @PaulSinclair: Imagine drawing an equilateral triangle on the surface of a sphere such that one point is at the north pole and the other two are on the equator. If you make the construction shown here on the sphere -- connecting the bisection points, then the angles at the bisection points add to 180 but the sum of the corners of the big triangle is 270. Plainly something is wrong. If we examine the postulates, the one we have violated is the parallel postulate. The problem is more specifically that this construction does not produce similar triangles in a world without it. – Eric Lippert Apr 09 '18 at 17:24
  • @PaulSinclair The existence of a pair of similar but not congruent triangles is yet another assertion equivalent to the parallel postulate. – Ethan Bolker Apr 09 '18 at 17:53
  • @EthanBolker - it was obviously dependent on the parallel postulate (which for Euclid is actually the statement that the sum of the angles of a triangle is $180^\circ$, after all). As I said, I was shy on time to track it down. I originally thought that the verification the triangles were congruent was being done by SSS - but I note now that this only applies to the center triangle. The others indeed depend on being similar to the main triangle, but each with a half-length side. – Paul Sinclair Apr 09 '18 at 20:52
  • -1 for the animation which plays automatically and which the user cannot control. If you want to include an animation, make so that it is initially paused, with a play/pause toggle. In any case, the demonstration didn't need animation, as other users, who have posted similar answers, have shown. – Rosie F Feb 17 '20 at 20:51
19

Draw a circle around a triangle. The central angle of each vertex is twice its size and the three central angles make the full round, that is $360$. Hence, the sum of interior angles of the triangle is one half of $360$, that is $180$.

Glorfindel
  • 3,929
  • 10
  • 24
  • 36
farruhota
  • 30,556
  • 2
  • 16
  • 49
  • 1
    A good answer, so upvoted. But the fact that the central angle is twice the angle formed by the chords has for 65 years felt unintuitive to me. I always have to check the proof (mentally) - which depends essentially on the theorems in Euclid that also lead to the theorem about the sum of the angles of a triangle. – Ethan Bolker Apr 08 '18 at 14:14
  • 2
    +1. The Inscribed Angle Theorem provides one of my favorite *affirmations* of the Angle-Sum Theorem (in the same way as you have described it), and, as such, would seem to help OP "see" the AST. However, the question also seems to be about *proof*. Since the standard proofs of the IAT *rely on* the AST, the IAT cannot be used to justify the AST. I wonder: Is there a neat proof of the Inscribed Angle Theorem that side-steps the Angle Sum Theorem? – Blue Apr 08 '18 at 14:15
  • @Blue I'm quite sure the IAT is equivalent to the AST, which is in turn one of the many equivalent forms of the parallel postulate. It would indeed be nice to see it proved directly from that postulate, and conversely. – Ethan Bolker Apr 08 '18 at 14:26
  • 1
    @EthanBolker: Agreed. IAT is *certainly* equivalent to AST (and, thus, Euclidean Parallel Postulate). For the benefit of readers who may not know: Triangle angle sums are *(almost-)anything but* $180^\circ$ in non-Euclidean geometry, yet the corresponding "central angle sum" remains $360^\circ$. So, the insight of the above answer only goes so far. – Blue Apr 08 '18 at 14:42
  • This proof should distinguish between acute and obtuse triangles. – user133281 Apr 08 '18 at 18:58
  • @user133281 I don't think that distinction is necessary. The circumcircle for a triangle with an obtuse angle works just fine. – Ethan Bolker Apr 09 '18 at 00:09
  • 1
    @Blue OP here, just wanted to add that this question was inspired by wanting a better understanding of thr Inscribed Angle Theorem . – hyperpallium Apr 09 '18 at 01:26
  • @user133281 The inscribed angle theorem is also true for obtuse angles (I also didn't think so originally, but its proof works for obtuse angles - try making the inscribed arc larger and larger, moving the arms of the angle closer and closer to the its vertex). – hyperpallium Apr 09 '18 at 01:35
  • I'm not sure this proof doesn't assume what it wants to explain. How would you prove the theorem that the central angle is twice the angle subtended at the circumference, for example? Somehow, the angle sum theorem feels very fundamental and related to the fact that we're doing **plane** geometry here. – Allawonder Apr 10 '18 at 15:30
17

Not a proof, but a nice way to "see it" dynamically: extend the sides of the triangle so you can see their exterior angles.

Exterior angles of a triangle

Now "zoom out" really far away from the picture. The triangle will look like a point, and angles $1$, $2$, and $3$ will have to sum to $360$. However, since these together with the interior angles sum to $3\cdot 180$, the interior angles themselves must sum to $180$.

You could also think of the angles sliding along each other similar to how a camera shutter closes.

Steven Gubkin
  • 7,856
  • 22
  • 46
  • I'm finding this to be very difficult to understand. I'm not sure _why_ angles 1, 2 and 3 have to sum to 360, or _why_ those angles together with the interior angles sum to 3 * 180. This seems to be only useful to someone that already knows a lot about triangles. – Clonkex Apr 09 '18 at 22:38
  • 1
    @Clonkex If you imagine zooming out, the three angles are just rays emanating from a single point: hence sum to 360. The exterior and interior angles sum to 3*180 because they together make three straight lines. – Steven Gubkin Apr 09 '18 at 23:59
  • This seems like a very roundabout way of demonstrating this, and doesn't actually show that the interior angles sum to 180. The lines emanating from the sides show (in an awkward kind of way) that the exterior angles sum to 360. However that doesn't show anything about the interior angles. You're saying `(exteriorAngles + interiorAngles) - exteriorAngles == interiorAngles`, which is... well, pointless. As far as I can tell, the only thing your answer does is show that the exterior angles sum to 360. Please correct me if you think I'm wrong. – Clonkex Apr 10 '18 at 04:11
  • 1
    @Clonkex I think you missed this line of the explanation: 'However, since these together with the interior angles sum to $3\cdot180$...' His explanation is perfectly OK. – Allawonder Apr 10 '18 at 08:22
  • @Allawonder No no no, that's exactly my point. It never says how you find the interior angles, therefore it's pointless for showing that the interior angles will be sum to 180. Is it really that hard to see what I mean? – Clonkex Apr 10 '18 at 11:53
  • @Clonkex Exterior angles = 2*180. (Exterior angles + interior angles) = 3*180 because the exterior angles AND the interior angles together form three straight lines, which are each 180. So Interior angles = (Exterior+Interior) - Exterior = 3*180-2*180=180. – Steven Gubkin Apr 10 '18 at 13:26
  • @Clonkex This answer is similar to "`the exterior angles were superimposed, to show they summed to 360`' in the question. The algebra following that in the question shows another sequence to get from the exterior angle sum to the interior angle sum. It's a little abbreviated, and uses the fact that a pair of interior and exterior angles sum to 180 (because they are on a line i.e. are supplementary). – hyperpallium Apr 10 '18 at 14:12
  • @Clonkex But it's obvious. The angles at a point on a straight line sum to $\pi$. – Allawonder Apr 10 '18 at 15:08
14

If you draw the triangle on a piece of paper, then cut off each corner, you can rearrange the corners so that the angles add up to a straight line.

Here is my simplest visual explanation courtesy of Paint:

enter image description here

lioness99a
  • 4,955
  • 2
  • 11
  • 28
  • In terms of actual ability to _see_ how the internal angles of a triangle add up to 180°, this is the best answer. I don't know why it had a downvote. Though I would say that step 4 is unnecessary and doesn't add anything useful. By step 3 you can already see that the angles make up to 180°. – Clonkex Apr 09 '18 at 01:29
  • 3
    @Clonkex I think this method is not "strong" enough. Cutting and moving and rotating corner pieces in imagination without changing the angle is hard. It may not be that obvious for someone who doesn't think of this 180 degree theorem as a fact. – AHB Apr 09 '18 at 11:33
  • @AHB As someone who didn't realise all internal angles of a triangle equal 180 was a thing, this was by far the most clear explanation. Also, I would expect anyone that's unsure would just get an actual piece of paper and some scissors and physically cut out the angles to prove it. I found that all the other explanations were extremely overcomplicated and didn't seem to prove anything. This one is the most visual of all. – Clonkex Apr 09 '18 at 11:38
  • 6
    @Clonkex This seems to work well for any individual triangle, but it's a lot less clear how to take this example-driven knowledge and learn something about *all* triangles (esp. compared to other answers, which make the generalization step quite explicit). That is, with this "proof", I would perpetually worry that, due to a failure of my imagination, I simply had not yet thought up a particularly wonky triangle that, when I cut it up and rearranged it, would *not* give a straight line. – Daniel Wagner Apr 09 '18 at 11:50
  • @DanielWagner I'm very confused by that. All these examples show the _same thing_, only this one does it in a simpler way. The most upvoted answer by Ethan Bolker, for instance, shows _exactly_ the same thing as this answer, except that I couldn't understand what it was meant to demonstrate until I saw this answer. Maybe it's different for people that enjoy maths. Personally I'm not a fan and very simple, real-world and practical proofs like this one are by far and away the easiest way for me to grasp a concept. – Clonkex Apr 09 '18 at 22:27
  • @Clonkex Upvoted because it demonstrates the concept, and makes a plausible case. But "proof" is a technical term in mathematics, meaning given accepted assumptions, and using accepted reasoning steps, you can get to the result - like a syllogism (or chess, with starting positions and movement rules). It matters because: how can you be sure the angles form a line *exactly*? You could magnify til you get to atoms... then what? A proof is better than reality. (But not *absolute* truth - how can we be sure the accepted assumptions and steps are true? At least we corral the doubt into them). – hyperpallium Apr 10 '18 at 01:28
  • @hyperpallium Yeah I'm aware of the true meaning of "proof", but think about it this way. Ethan Bolker's answer (which, mysteriously, is highly upvoted) shows _exactly_ the same thing as this answer, except that instead of trying to show that the line is perfectly straight it tries to show that the angles fit against a line. Either way relies on measuring angles (either the angles next to the line or angle of the line itself), and in either case if one is true (eg. the line is straight) then _the other must also be true_ (the angles add to 180), based on accepted assumptions. See what I mean? – Clonkex Apr 10 '18 at 03:56
  • @Clonkex Looking again at EthanBolker's answer, I see that none of the assumptions and steps are mentioned; I had filled them i unconsciously (because I just complete the 'High School Geometry" exercises and proofs at Khan Academy). I've often complained about gaps in math, but did 't see it here - I guess that's how gaps happen. – hyperpallium Apr 10 '18 at 13:25
  • 1
    @Clonkex Ethan's answer adds a parallel line. It is accepted that *alternate interior angles* are the same (that is, when a another line crosses (*transverses*) parallel lines, "interior" angles are on the sides between the parallel lines; and "alternate" means on opposite sides of the transverse line.) In the diagram, the two blue angles are alternate interior angles, and the two red angles are alternate interior angles. These angles sum to 180 (*supplementary*, done in pairs). So there's no actual measuring; the proof follows from accepted assumptions. – hyperpallium Apr 10 '18 at 13:35
  • @Clonkex I think the point here is that Ethan Bolker's answer shows how one can cut out (abstractly) the angles without changing them by those parallel lines. I regard this answer as a motivation to Bolker's, for first timers. This shows where the idea of that proof comes from, but it's not very airtight, when one thinks about it. – Allawonder Apr 10 '18 at 15:27
  • @DanielWagner (and others) This cut and rearrange visualization is convincing, but there's a subtle missing point. If you took a triangle on a sphere - think a piece of the peel of an orange whose edges are great circles - cut out the angles and rearranged them around a common vertex on the sphere you'd get more than a straight angle. In fact it's not hard to see a spherical triangle with three right angles. That fact that Euclidean triangles are in a plane where parallel lines exist is crucial to the argument. – Ethan Bolker Apr 11 '18 at 00:26
7

In a similar vein to your walking around the exterior and obtaining an external angle sum of 360 degrees, you can do the same by "walking" inside the triangle. The demonstration is best with physical barriers for sides.

Make a triangle using for example, blocks of wood or standing cardboard. Place a pen inside one corner against one side. Record the pen's orientation. Slide it along the side it is touching, and when you reach the end, swing it inside the triangle to rest against the other side. Repeat until the pen is back in the same position it started.

Observe that the pen has been spun exactly halfway around by spinning inside the angles of the triangle, so together the angles of the triangle must be 180 degrees.

This process can be extended to demonstrate physically the internal angle sum of any shape with two or more sides - for two sides the angle is zero as the pen cannot spin at all and for quadrilaterals or higher the angle is given by the usual formula, observed as one complete spin for every 360 degrees.

Nij
  • 2,841
  • 3
  • 22
  • 31
  • 1
    'Repeat until the pen is back in the same position it started.' I am thinking that if I return back to the direction I was facing, then I must have gone a full turn. Half a turn would be directly opposite my original direction. Your demonstration only shows why the exterior angles sum to a full turn. Draw it and see. – Allawonder Apr 09 '18 at 21:09
  • 2
    @Allawonder I also thought this, but it's just worded poorly. What Nij means is actually not to "drive" the pen around as though it were a car, but rather when you reach each corner of the triangle to rotate the pen around the point closest to the corner, such that it then points (with the other end) towards the next corner. So the pen is effectively doing a sort of 3-point turn. Drive forward, rotate slightly, drive backward, rotate again, drive forward, rotate once more. At that point it'll be 180 backwards to how you started. Not the most obvious proof/example IMO but I guess it works. – Clonkex Apr 09 '18 at 22:34
  • The same **position**. *Not* the same **orientation**. The words were chosen for a reason. It's very clearly an action within the triangle and leads to exactly half a turn, which suggests a reader not paying attention. – Nij Apr 09 '18 at 23:59
  • 2
    @Nij I think I missed this part last night: '...when you reach the end, swing it inside the triangle to rest against the other side.' Works quite alright, but I can't say why I feel dissatisfied about it. – Allawonder Apr 10 '18 at 08:34
  • 1
    @Clonkex Yes, I eventually got it. I agree it's not a good enough description. – Allawonder Apr 10 '18 at 08:37
  • 1
    @Nij Uh... saying **position** does not automatically imply **not orientation**. Just because you say the pen reached the _position_ it started in does not mean you don't also expect it to have reached the _orientation_ in which it began. The most obvious thing to do is drive the pen around like a car, always moving forwards, and in that situation the pen will rotate 360°. It simply seemed like you somehow thought the pen would end up only rotating 180°, hence our confusion. Not sure what you mean by "It's very clearly an action within the triangle" though... – Clonkex Apr 10 '18 at 12:01
  • I've come to like this after fooling around with it a bit, but I reimagined it my way. Not comfortable with a pen, I imagined myself walking along the walls of a room with a polygonal plan. So for a triangle, if I start at one side facing the wall and walk sideways always, it becomes clear that the only way to turn at the corners is by unfacing the wall. At every corner this sequence is alternated. So if I denote facing the wall by a + and facing the room by a -, then if I start out facing a wall, I turn at one corner to face the room -, then at the second corner face the wall again +, then... – Allawonder Apr 10 '18 at 15:14
  • ...finally face the room, -. This is directly opposite to how I started out. Thus I can conclude with certainty -- not that I have made a total turn of $\pi$ -- but that I have made a total turn of a multiple thereof. This becomes clear when you use a pentagon instead. You make three about-turns (draw lines parallel to one side and passing through the polygon through the vertices). For polygons with an even number of sides, you always end up facing the same direction, so that we conclude that we've turned a total of $2\pi n$. I'm still trying to see how this connects to the parallel postulate. – Allawonder Apr 10 '18 at 15:18
  • That would be "obvious" if it was following the instruction to swing the pen inside the triangle, from one side to the next, inside the corner. But it isn't. Please stop making comments that seem to imply a lack of comprehension that trends towards deliberate. – Nij Apr 10 '18 at 19:17
4

It isn't a proof, but if you fold the bottom two angles in, then fold the angle opposite the base down, it will just fit together with the other two to form a straight angle. Try it with a piece of paper. Rather nifty.

The proof I know and like is the one in Ethan Bolker's answer.

  • 1
    "Fold in" means (reflect) along a perpendicular to the base, so the folded base aligns with itself? And, I think you'd need to fold the top angle first (reflect along a parallel to the base), so you'd know how much to fold the bottom edges in? – hyperpallium Apr 09 '18 at 01:24
  • 1
    @hyperpallium yes; I guess you could fold the base angles in too far, or too little, otherwise –  Apr 09 '18 at 01:29
  • Do the angles always exactly matxh up? If one could show this, it *would* be a proof (but might lose its simplicity). A line through the origin $y=mx$ reflected across the x-axis becomes $y=-mx$. If reflected across the y-axis, it also becomes $y=m\cdot(-x)=-mx$. So the folded angles have the same slopes (are parallel), so do match up. Not sure how to express all that visualy/geometrically... – hyperpallium Apr 10 '18 at 23:37
  • 1
    @hyperpallium They must. Tempting to call it a proof... –  Apr 11 '18 at 00:11
  • I don't think it can be folded for an obtuse base angle. However, one can just rotate the obtuse angle to the apex. Assuming there's only one obtuse angle (which might **not** be a reasonable assumption, considering what's being proved...) – hyperpallium Jul 19 '18 at 02:20
  • It seems like a reasonable assumption, as it's fairly obvious, to me, that with two obtuse angles you don't get a triangle... –  Jul 19 '18 at 02:43
  • I just meant that sum to 180 implies we can't have two obtuse angles. To assume that is begging the question. – hyperpallium Jul 19 '18 at 03:49
4

Euclid Elements, book I, proposition 32

Elememts, I. 32

I've reflected it, and used colours to try to to match the top answer here.

Though asymmetric, it's useful in proving other theorems, as it also shows: an exterior angle equals the sum of the opposite interior angles (blue and red).

hyperpallium
  • 953
  • 7
  • 16
2

Every triangle could be inscribed in a circle.

Each inscribed angle measures half the arc opposite to it.

The angle sum of the triangle inscribed in a circle is half the total arc measures around the circle.

The total arc measures around a circle is $360$ degrees.

Thus the angle sum of a triangle is $180$

Mohammad Riazi-Kermani
  • 67,745
  • 4
  • 36
  • 86
1

Imagine a triangle $ABC$ with two of the sides, say $AB$ and $BC$, rigid but meeting at the joint $B$, which is movable in the plane of the triangle, like a hinge sort of (also, you may fix one of these rigid sides, say $BC$, and rotate the other). Let the side $AC$ be made of a sufficiently elastic material (at least so that $AC\leq AB+BC$ ).

If you play with this imaginary object, you should soon see that as you increase $A \hat B C$, the other two angles eventually diminish (certainly for $A \hat B C>π/2$). Indeed, you may extend $\angle ABC$ as wide as you please close to $\pi$, so that the two rigid sides approach a single rectilinear segment; then the angle between them becomes arbitrarily close to $π$ and the other two angles are vanishing; in the limiting (degenerate) position the side $AC$ coincides with the segment formed by the rigid sides.

If you can already see that any point in this sequence of transformations of $\triangle ABC$ the sum of angles is conserved, then it is clear why for any triangle this sum is always $π$.

Allawonder
  • 12,848
  • 1
  • 15
  • 26
  • 1
    I find this kind of approach the most appealing, but it's difficult to read (I actually deferred reading it at first because it's intimidating). I think it's due to its formal detail and radians (degrees are standard for geometry, aren't they?). I like that one can move the hinge arms independently. However, I think that the change in vertex angle (at $B$) and change in another angle compensate each other (i.e. have constant sum) requires proof - which could use the alternate interior angles of @EthanBolker's answer, with a line parallel to $AC$ (also the line when $B=180$). – hyperpallium Apr 09 '18 at 01:17
  • 1
    @hyperpallium Done some editing. Should be clearer now. And yes your observation is true. This argument assumes that one already sees that the angle sum is constant; it only confirms that the value of this conserved quantity is $\pi$. – Allawonder Apr 09 '18 at 21:46
  • To this end I've written another answer that takes better care of this. – Allawonder Apr 09 '18 at 21:50
1

This explanation makes it clear why this particular theorem about triangles in the plane is deeply connected to the parallel postulate.

Consider two parallel lines. Lay two parallel transversals along them, so that the four lines determine a parallelogram. Then by the parallel postulate the two interior angles cut off on the same side by a transversal falling on two parallel lines are supplementary -- that is, they add up to two right angles. Consequently, the sum of angles in any parallelogram is four right angles, or $2π$.

Now it is clear that a parallelogram can be decomposed (along a diagonal) into two congruent triangles; and conversely, that two congruent triangles, plus a finite sequence of euclidean transformations, can be made to determine a parallelogram. In other words, two copies of a triangle determine a parallelogram. Consequently, the sum of angles in any triangle is half that of the related parallelogram, which is always $2π$. The result follows.

Edit. An even shorter way to see the connection to the parallel postulate is this. Let a straight line fall on two straight lines so that the two interior angles on one side of the transversal are less than supplementary; we have three lines here. Keep the transversal and either of the other two lines fixed, then rotate the last about its point of intersection with the transversal so that the two angles in question gradually approach supplementarity. Now, at all times when they're less than supplementary, we always have a triangle on that side (by the postulate). As you make the angles closer to supplementary, the distance from the transversal to the point where the lines meet increases indefinitely (and the angle at that meeting point becomes smaller and smaller). In the limit when the lines are parallel (i.e., they don't meet anywhere), the angles in question are just supplementary.

Allawonder
  • 12,848
  • 1
  • 15
  • 26
  • This one sort of sneaks up on you, then pow! I wondered about this equivalence since @EthanBolker. The "parallel postulate" is alternate interior angles? I think "flipped" is ambiguous, and (to me) its stronger meaning is "reflected" (rotated about a line in the plane, in 3D), but here you mean "rotated" (within the plane). Finally, is the triangle sum half the parallelogram sum simply because the two triangles' angles must sum to the parallelogram's, and the triangless' are equal through congruency? [nvm, you start with two copies, so of course they sum to the parallelogram's. nice] – hyperpallium Apr 10 '18 at 01:49
  • @hyperpallium #1. No, the parallel postulate is this: If a straight line (transversal) falls on two straight lines so that it makes the two interior angles on one side less than two right angles, then the lines meet on that side. Obviously, this means that if those two angles are just supplementary, then the lines are parallel. #2. No, I obviously didn't mean rotation in the plane. Flipping a triangle about one of its sides can have one and only one meaning -- it is perfectly clear (I think you did not pay attention to the axis of flipping here); and no, reflection is just synonymous to... – Allawonder Apr 10 '18 at 07:58
  • ...flipping as used here. Indeed, I used that word instead because of your complaint about inaccessibility and also -- I'm only making you **see** why this is true; it's not a proof, strictly speaking. #3. Since the triangles are congruent -- by a finite sequence of euclidean transformations this is clear -- the sum of angles in one is exactly half the sum of two. – Allawonder Apr 10 '18 at 08:01
  • I've added another way to see the connexion to the parallel postulate. It came to mind while clarifying the points you asked for. – Allawonder Apr 10 '18 at 08:12
  • Sorry I came across as complaining; I was trying to be helpful, but it's a fine line. I felt a little hesitant commenting. Maybe it can't be made much simpler without losing precision. Ethan Bolker's diagram is simple - but rests heavily on background knowledge. "Flipped about one side" (to me) means the *side* is the axis of rotation. "Flipping (rotating) about the midpoint of the side" is how I'd say it. But I can see your interpretation. I'll have to think about the parallel postualate connection - it's a bit over my head! Thanks for the responses and edits! – hyperpallium Apr 10 '18 at 13:58
  • @hyperpallium '"Flipped about one side" (to me) means the *side* is the axis of rotation. "Flipping (rotating) about the midpoint of the side" is how I'd say it.' Yes, I mean flipped about one side, like when you open turn over a leaf of a book. The leaf is flipped, as it were, about the spine of the book. – Allawonder Apr 10 '18 at 15:03
  • To clarify here, this flipping transcends the plane. (You're right, one can prove congruency by using only combinations of translations and rotations, but I think reflecting does it most naturally). – Allawonder Apr 10 '18 at 15:05
  • But if you flip about one side (like `the spine of the book`), you get a kite, not a parallelogram. BTW backticks are a way to quote in comments – hyperpallium Apr 10 '18 at 23:27
  • 1
    @hyperpallium Oops, my bad! That's true. Imagine I've been mentally deceived since into assuming that a pure flip does it. Little wonder the original confusion. I guess I have to apologise here for drawing on something that was a no brainer -- I wasn't actually trying it; I rested on the false security that it was so. Hasty conclusion on my part. Sorry! I'll change the answer accordingly, too. Thanks. – Allawonder Apr 11 '18 at 06:07
  • 1
    No worries! I had to check a diagram (I used Rodrigo A. Pérez's answer, the current second one, with a tiling of parallelograms) to see the effects of reflection vs. rotation. Anyway, to reiterate, i like your novel approach the two triangles' sum equalling the parallelograms' sum. – hyperpallium Apr 11 '18 at 08:04
1

If you accept that the sum $S$ of angles is the same in every triangle there’s another proof. Choose a point $D$ on $AB$. Now the sum of angles in the triangles $ADC$ and $DBC$ is $2S$:

$$2S=A+\angle ADC+\angle DCA+B+\angle DBC+\angle BCD=A+B+C+180,$$ hence $S=180$.

Michael Hoppe
  • 16,684
  • 3
  • 28
  • 47