# A way to directly see that the interior angles of triangle sum to $180^\circ$?

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...?
• The typical proof introduces a line through a vertex, parallel to the opposite side. – Blue Apr 8 '18 at 12:39
• Draw a parallel line (how?) and consider internal alternate angles. – Piquito Apr 8 '18 at 12:42
• 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 8 '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 9 '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. This is the same as many of the answers, just done with a piece of paper. – N. Owad Apr 12 '18 at 1:28

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:

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"

• I think this answer is better than mine. – Ethan Bolker Apr 8 '18 at 15:21
• Hate to be that guy, but I can't see it. Could someone include an image? – user1717828 Apr 9 '18 at 1:45
• @user1717828 Look at this. 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 $a+b+c=180$. – Billy Rubina Apr 9 '18 at 2:29
• @user1717828 I added an image. The poster may remove it if he wants to leave it to your imagination. – Ethan Bolker Apr 9 '18 at 14:25
• @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 9 '18 at 21:56

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.

• 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 9 '18 at 12:32
• very beautiful and understandable. – Mohammad Riazi-Kermani Apr 9 '18 at 13:32
• @PaulSinclair SAS similarity is equivalent to the parallel postulate. See Petrunin's notes on geometry for example. – Steven Gubkin Apr 9 '18 at 13:52
• @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 9 '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 9 '18 at 17:53

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 $180$.

• 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 8 '18 at 14:14
• +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 8 '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 8 '18 at 14:26
• @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 8 '18 at 14:42
• @Blue OP here, just wanted to add that this question was inspired by wanting a better understanding of thr Inscribed Angle Theorem . – hyperpallium Apr 9 '18 at 1:26

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:

• 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 9 '18 at 1:29
• @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 9 '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 9 '18 at 11:38
• @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 9 '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 9 '18 at 22:27

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

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.

• 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 9 '18 at 22:38
• @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 9 '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 4:11
• @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 8:22
• @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

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.

• '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 9 '18 at 21:09
• @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 9 '18 at 22:34
• @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 8:34
• @Clonkex Yes, I eventually got it. I agree it's not a good enough description. – Allawonder Apr 10 '18 at 8:37
• @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

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.

• "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 9 '18 at 1:24
• @hyperpallium yes; I guess you could fold the base angles in too far, or too little, otherwise – Chris Custer Apr 9 '18 at 1:29
• 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
• @hyperpallium They must. Tempting to call it a proof... – Chris Custer Apr 11 '18 at 0:11
• 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 2:20

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 $180$

• 'The angle sum of the triangle inscribed in a circle is half the total arc measures around the circle.' What does this mean? – Allawonder Apr 10 '18 at 8:26
• Which part of the sentence you do not get? – Mohammad Riazi-Kermani Apr 10 '18 at 8:38
• I mean, you state this as some axiom or commonly accepted fact, but unless we're using words differently I don't see how this could be so. – Allawonder Apr 10 '18 at 8:50
• Thanks, I edited my solution just now to address your concern. – Mohammad Riazi-Kermani Apr 10 '18 at 8:54
• 'Each inscribed angle measures half the arc opposite to it.' Why so? – Allawonder Apr 10 '18 at 9:16

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).

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 $π$.

• 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 9 '18 at 1:17
• @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 9 '18 at 21:46
• To this end I've written another answer that takes better care of this. – Allawonder Apr 9 '18 at 21:50

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.

• 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 1:49
• @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 7:58
• ...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 8:01
• @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 6:07
• 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 8:04

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 $S=180$.

## protected by J. M. is a poor mathematicianApr 9 '18 at 16:38

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