Back in grade school, I had a solution involving "folding the triangle" into a rectangle half the area, and seeing that all the angles met at a point:

triangle proof

However, now that I'm in university, I'm not convinced that this proof is the best one (although it's still my favourite non-rigorous demonstration). Is there a proof in, say, linear algebra, that the sum of the angles of a triangle is 180 degrees? Or any other Euclidean proofs that I'm not aware of?

  • 16
    $\begingroup$ You need some version of the parallel postulate to prove this. One version takes precisely this-"the angles of a triangle add to $180^\circ$". This is equivalent to the parallel postulate. $\endgroup$ – Ross Millikan Jan 3 '13 at 18:51
  • $\begingroup$ Hmm, in grade school, we took the parallel postulates independently of the sum-of-the-angles-of-a-triangle postulate. How would you prove the parallel postulates from the sum of the angles of a triangle being 180 degrees? (I know how to do it the other way; one of the answers down there was one I found online while researching this problem.) $\endgroup$ – Joe Z. Jan 3 '13 at 18:59
  • 2
    $\begingroup$ I would think, given a line and a point not on the line, you drop a perpendicular from the point to the line, then construct a line a right angles to the perpendicular. If the two lines meet on either side, you have a triangle with more than $180^\circ$, so the lines don't meet and you have constructed a parallel. $\endgroup$ – Ross Millikan Jan 3 '13 at 19:22
  • 4
    $\begingroup$ Maybe MSE: sum-of-angles-in-a-triangle. Also, Girls Get Curves $\endgroup$ – Amzoti Jan 3 '13 at 20:01
  • 1
    $\begingroup$ Did anybody ever tell you about a triangle with 3 90 degree angles on it? (Seriously!) $\endgroup$ – jcolebrand Jan 5 '13 at 9:28

10 Answers 10


According to the Gauss-Bonnet theorem, if $T$ is your triangle, $\gamma_i$ its sides and $v_i$ its vertices, $$\int_T K+\sum_i\int_{\gamma_i}\kappa + \sum_i\alpha_i=2\pi\chi(T)$$ with $K$ the Gaussian curvature, $\kappa$ the geodesic curvature along the sides, $\alpha_i$ the external angle at the vertex $v_i$ (measured in radians), and $\chi(T)$ the Euler characteristic of $T$. Since the plane is flat, $K\equiv 0$; since the sides of the triangle are geodesics, $\kappa\equiv0$; and since $T$ is contractible, $\chi(T)=1$. Therefore the above formula tells us that $$\sum_i\alpha_i=2\pi.$$ Since the internal angle at the $i$th vertex is $\pi-\alpha_i$, this tells us that the sum of the internal angles is $\pi$, which is what we wanted

| cite | improve this answer | |
  • 34
    $\begingroup$ (I am not serious, of course...) $\endgroup$ – Mariano Suárez-Álvarez Jan 3 '13 at 18:36
  • 11
    $\begingroup$ Yes, it is perfectly valid. $\endgroup$ – Mariano Suárez-Álvarez Jan 3 '13 at 18:58
  • 89
    $\begingroup$ Peanut, meet Sledgehammer. Hi, Sledgehammer. $\endgroup$ – AndrewG Jan 3 '13 at 18:58
  • 8
    $\begingroup$ +1 As the "best" proof, in the sense that it will fully generalize to triangles on any 2-manifold $\endgroup$ – user7530 Jan 3 '13 at 19:17
  • 8
    $\begingroup$ @user7530, in a sense, the Gauss-Bonnet theorem is the generalization. $\endgroup$ – Mariano Suárez-Álvarez Jan 3 '13 at 20:05

Here's a decent Euclidean proof:

Let $x$ be the line parallel to side $AB$ of $\triangle ABC$ that goes through point $C$ (the line is unique because of the fifth postulate). $AC$ cuts $x$ and $AB$ at the same angle, $\angle BAC$ (corollary of the fifth postulate). $BC$ cuts $x$ and $AB$ at the same angle, $\angle ABC$. These two angles and the final angle $\angle ACB$ form a straight angle on $x$, which is always $180^\circ$ (corollary of the third postulate).

enter image description here

| cite | improve this answer | |
  • 3
    $\begingroup$ That's what we learned at school. $\endgroup$ – Nakilon Jan 4 '13 at 13:51
  • 2
    $\begingroup$ This proof seems more or less identical to the one you present in your question. $\endgroup$ – Greg L Jan 4 '13 at 14:32
  • 1
    $\begingroup$ +1 this proof is the closest to the "it's just intuitively obvious" answer that I wanted to post. :) $\endgroup$ – Robin Winslow Jan 4 '13 at 16:51
  • 3
    $\begingroup$ Beautiful. Just to emphasize what may be obvious to most, since this depends on the fifth postulate it does not hold true in non-Euclidean geometries. In spherical geometry for instance the sum of the interior angles of a triangle will always exceed 180. $\endgroup$ – TimothyAWiseman Jan 4 '13 at 17:45
  • 1
    $\begingroup$ @ Joe Zeng: Yes but when you "fold" the triangle you are just reflecting the top vertex of the triangle onto another line parallel to the base of the triangle (but that line just happens to be the base itself, i.e. a line is parallel to itself). The reflection preserves angles and the final geometric analysis is similar (just rotated 90 degrees). I'm not disagreeing with you per se, just pointing out how your "folding" technique is actually more rigorous than you give it credit for! $\endgroup$ – Greg L Jan 4 '13 at 21:06

There is always the "taxicab" proof. Start at a vertex and take a cab trip around the triangle. The cab will have turned through an angle of $360^{\text{o}}$ Extend the sides of the triangle appropriately and you will get three $180^{\text{o}}$ angles with the vertex angles being supplementary to the turns taken by the cab. Then note that $540-360=180$ ...

I know this isn't a direct answer, but it is my favourite demonstration.

| cite | improve this answer | |

This demonstration adopts the point of view that Euclidean geometry is the study of the affine group in the plane (generated by reflections in $\mathbb{R}^2$). This group includes all translations and rotations. They preserve distances and angles. The role of this group, and in particular how parallel translation of vectors interacts with rotation, is prominent in the argument.

Translating triangle $ABC$ by vector $AB$ produces triangle $A'B'C'$ (which establishes that angle $ABB'$ is straight) and rotating $ABC$ around the midpoint $M$ of $BC$ produces triangle $A_1'B_1'C_1'$. Notice that $C=B_1'$ because $M$ is the midpoint of $BC$ and, also by construction, $C_1' = A' = B$.


Everything hinges on $A_1'$ coinciding with $C'$, as suggested in this figure. Although this is visually "obvious," there is something to be shown, because $A_1'$ is the result of rotating $ABC$ while $C'$ is the result of translating $ABC$. Among the many ways to make this demonstration, then, here is one that is geometric in spirit.

The image of vector $AB$ under the ($180^\circ$) rotation about $M$ is the vector $A_1'B_1'$. Because $180^\circ$ rotations negate vectors, $A_1'$ must be the translate of $B_1'$ by $AB$. Since (by construction) $C'$ is the translate of $C = B_1'$, $C'$ and $A_1'$ are coincident.

This shows that the straight angle $ABB'$ is decomposed into non-overlapping angles $\beta$, $\gamma'$, and $\alpha'$. Because translations and rotations preserve angles, the measure of angle $\alpha$ equals the measure of $\alpha'$ (its image under the translation) and the measure of angle $\gamma$ equals the measure of $\gamma'$ (its image under the rotation), whence the sum of the measures of $\alpha$, $\beta$, and $\gamma$ is the measure of a straight angle, $180^\circ$.

| cite | improve this answer | |
  • 1
    $\begingroup$ I see that rschwieb recently sketched the same approach. The figure may help in understanding it and the appeal to properties of the affine group should help make it rigorous, especially insofar as a crucial gap--proving the coincidence of $A_1'$ and $C'$--is concerned. $\endgroup$ – whuber Jan 3 '13 at 23:05
  • $\begingroup$ What have you used to sketch this figure? $\endgroup$ – B Faley Jan 4 '13 at 8:26
  • 1
    $\begingroup$ @Meysam Geogebra is a free Java application designed for geometric illustrations. It's mature and works well. $\endgroup$ – whuber Jan 4 '13 at 13:53
  • $\begingroup$ @whuber Thanks for recognizing our solutions have the same idea. I'm glad you offered more details: it's not my field, and I was just providing the easiest explanation I could think of. $\endgroup$ – rschwieb Jan 4 '13 at 15:11

You always have the "original" proof found in Elementa. I like this one. It is direct and easy to follow, making only use of addition of angles and properties of parallell lines. It does resemble your proof, but replaces "folding" of the triangle by the construction of a line parallell to one of the sides.

| cite | improve this answer | |

There's nothing wrong with your "folding" proof. Fold = reflect. To formalize it, drop a perpendicular from the top vertex to the base. Call that point $P$. Then reflect the top vertex across the top horizontal side of the rectangle, and reflect each of the bottom vertices across the corresponding vertical side of the rectangle. Note that all 3 reflections coincide at $P$. End of proof.

An exercise: This result isn't true in hyperbolic or spherical geometry. Where does this proof use an axiom specific to Euclidean geometry?

| cite | improve this answer | |
  • $\begingroup$ I think maybe the definition of reflection would use the parallel postulate, but I'm not sure. $\endgroup$ – Joe Z. Jan 3 '13 at 19:19
  • 3
    $\begingroup$ Reflection itself is okay in other geometries, but the problem comes when you show that reflections coincide at $P$. Essentially, you need this result: Given a right triangle, if a perpendicular is drawn from the midpoint of the hypotenuse to one of the legs, then it hits the midpoint of the leg. To prove this, you need an argument with similar triangles or the like, which is only available in Euclidean geometry. $\endgroup$ – Ted Jan 3 '13 at 19:47

How about this:

Look at your triangle so that one side is horizontal. Duplicate it and slide the copy horizontally until the two triangles touch only at a single point. Make another copy of the original triangle and rotate it 180 degrees and slot it in the gap between the first two triangles.

What you wind up with is a trapezoid where the three angles adjacent to each other, additively giving a 180 degree angle.

It seems that since the whole thing is done with three congruent triangles, it should be easy to argue from elements.

| cite | improve this answer | |

It's a (trivial) calculation.


Let $A$, $B$, and $C$ be distinct points in the plane. Considering them as complex numbers, recall that the angle $\theta$ at a vertex like $ABC$ is the angular part $\theta$ of $(C-B)/(A-B) = r\exp(i\theta)$. It is written $\theta = \arg(r\exp(i\theta))$ and is defined only modulo $2\pi$, although conventionally a representative in the interval $(-\pi, \pi]$ is taken.

Observe that $\arg(z w) =\arg(z) + \arg(w)$ for any nonzero complex numbers $z$ and $w$: this follows immediately from the multiplicative property of the exponential, $\exp(i\theta)\exp(i\phi) = \exp(i(\theta + \phi))$.


The sum of the three angles at vertices $A$, $B$, and $C$ is obtained merely by rearranging the product of three numbers in the denominator of a fraction:

$$\eqalign{ \arg\frac{B-A}{C-A} + \arg\frac{C-B}{A-B}+\arg\frac{A-C}{B-C} &= \arg\left(\frac{B-A}{C-A}\frac{C-B}{A-B}\frac{A-C}{B-C}\right) \\ &= \arg\left(\frac{B-A}{A-B}\frac{C-B}{B-C}\frac{A-C}{C-A}\right) \\ &= \arg\left(-1^3\right) = \arg(-1) \\ &= \pi. }$$

Because the points are distinct, all three angles are strictly between $-\pi$ and $\pi$. The only way three of them can sum to $\pi$ (modulo $2\pi$) is for the sum actually to equal $-\pi$ or $\pi$, QED.


Both sums, $-\pi$ and $\pi$, are possible: in the former case, the triangle $ABC$ is traversed in the negative direction (with its interior to the right) while in the latter it is traversed positively.

This argument does not generalize (without taking special care) to four or more points, because the usual result found in elementary geometry expositions is not generally true (namely, that the sum of interior angles of an $n$-gon is $(n-2)\pi$). E.g., with four points we might care to show that the sum of angles in a 4-gon (a "shape with four sides") is $2\pi$. Consider the points at $1$, $i$, $-i$, and $-1$: in that order. Each of the angles, considered in isolation, is $\pi/4$, so their sum is $\pi$. We need to use correctly oriented angles, however; in traversing this 4-gon, two of the angles will equal $\pi/4$ and two will equal $-\pi/4$, summing to $0$. This is correct: the result is congruent to $2\pi$ modulo $2\pi$, as it should be, but it is not equal to $\pm 2\pi$.

| cite | improve this answer | |
  • $\begingroup$ @ your non-generalization argument: Are you sure you didn't mean $1, -1, i, -i$? Because as it is now, your shape is just a square. $\endgroup$ – Joe Z. Jan 4 '13 at 16:17
  • $\begingroup$ @Joe You are right: I meant $1,i,-i,-1$, in that order (although your construction works just as well). I'll fix it--and thanks for reading so carefully! $\endgroup$ – whuber Jan 4 '13 at 16:19
  • $\begingroup$ Also, I think it'll be equal to $2\pi$ if the clsoed curve doesn't intersect itself? $\endgroup$ – Joe Z. Jan 30 '15 at 18:44
  • $\begingroup$ @JoeZ. Or $-2\pi$: the complex arithmetic keeps track of angle orientation, too. $\endgroup$ – whuber Jan 30 '15 at 19:41

Assuming that the sum of all angles of polygon does add up to a particular constant depending on the no. of sides of the polygon.

Let us assume that the constant is 'k', if at the end of the proof we are able to find a value k for an arbitrary triangle, we can assume that k exists for all the triangles.

Assume an arbitrary triangle.

By gen. convention, each angle of triangle is named as ang.A , ang.B, ang.C, hence,

(1) ang.A + ang.B + ang.C = k

Now we all know that the triangle can be divided into 3 portions by lines joining each vertex and any point inside the triangle , (example centroid). let us name it Point P.

let The angle subtended by each side on the Point P be cen.A, cen.B, cen.C. Thus,

(2) - cen.A + cen.B + cen. C = 360deg.

Also the line joining the vertex and the Point P. of triangle will divide each angle in two parts, name each as

(3) ang.A=ang.A1 + ang.A2 and so on for ang.B and ang.C

thus for each of the smaller triangles we have following equations -

(4) ang.A1 + ang.B1 + cen.C = k

(5) ang.A2 + ang.C1 + cen.B = k

(6) ang.B2 + ang.C2 + cen.A = k

(4) + (5) + (6) will imply that

ang. A + ang.B + ang.C + cen.A + cen.B + cen. C = 3k

Now using (2) we have

ang. A + ang.B + ang.C + 360deg. = 3k

But by initial assumption and hence by (1) we have

k + 360deg. = 3k



I think this simple proof should suffice.

Or rather just cut the triangle any which way you want keeping the vertices intact. Arrange them with their pointed ends together, voila, you can see the angles must add up to 180deg.

Also why go for complicated proofs when we have simple ones. I guess, because we are Mathematecians! :)

| cite | improve this answer | |
  • $\begingroup$ It is interesting that the conclusion follows from Assumption (1). However, this requires proof. After all, it is not true in other closely related geometries (such as spherical or hyperbolic geometries). $\endgroup$ – whuber Jan 4 '13 at 14:23
  • $\begingroup$ "Also why go for complicated proofs when we have simple ones." The Gauss-Bonnet proof was pretty simple. $\endgroup$ – Joe Z. Jan 4 '13 at 22:57
  • $\begingroup$ Interesting argument, thanks for sharing it. Welcome to Math.SE! $\endgroup$ – user53153 Jan 5 '13 at 9:16
  • $\begingroup$ @whuber-Could you explain how spherical or hyperbolic geometry is similar to a polygon. Because to me a polygon is a closed two dimensional figure which has only straight line segments as its sides, and the end-points as vertices. Though I am not even aware of hyperbolic geometry except as in a conic section. I have mathematics knowledge only upto what you would expect from an average sophomore, 1st year. $\endgroup$ – user1729 Jan 7 '13 at 5:21
  • 1
    $\begingroup$ @JoeZeng - To me anything that has calculus in it is atleast a bit more complicated than algebra . :) $\endgroup$ – user1729 Jan 7 '13 at 5:23

If we have a Triangle with vertices A,B,C

As we can see in the Triangle, if a man starts from Vertices A and goes in the direction of the line extended from Vertices A, suddenly decides to go towards Vertices B, Meaning he has to turn in Anti-Clockwise Direction for Angle(PI - A). now he walks to Vertices B, then suddenly he wants to go towards C and turns in Anti Clockwise Direction for Angle(PI - B) and moves along C, finally he does the same and turn towards A with a Rotation of Angle(PI - C), So completes One Revolution or 2*PI Radians, or the sum of all the Three Rotations the man took to complete one Perimeter walk on the Triangle Walk or an equivalent of Angle(3*PI -(A+B+C)).

Meaning we Reach an Equation : Angle(2*PI) = Angle(3*PI -(A+B+C)) => Angle(A+B+C) = Angle(2*PI)

Assumptions Taken to Calculate This result :

  1. A rotation of Angle(PI) is required to move in an exactly opposite direction.
  2. Obviously to come to the same line of sight we need to rotate an angle of Angle(2*PI).

I think i can safely assume these and get to the result.

| cite | improve this answer | |

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