# Internal angle sum of a triangle

I teach a 5th grade class geometry, and I came up with the following alternative proof (?) to show that the internal angle sum of a triangle is $$180^\circ$$. I remember reading that this result is equivalent to the parallel postulate, however I can't see where I have used this axiom.

I would be very happy if anyone could point out my mistake. Here we go:

The pencil starts at the side $$CB$$

The pencil turns clockwise $$\angle B$$

The pencil turns clockwise $$\angle A$$

The pencil turns clockwise $$\angle C$$. Since the pencil points in the opposite direction it has turned $$180^\circ$$ (Or maybe $$180^\circ+360^\circ\cdot k$$?)

• Interesting question! I suspect the parallel postulate comes into play when summing angles at distant locations - that seems to call for parallel translation. Imagine your argument on the sphere, with the big triangle that has three right angles. Mar 4 at 16:56
• If you have rotated the pencil about B by $\angle B$, then about $A$ by $\angle A$, have you really rotated the pencil by angle of $\angle A+\angle B$? If so, about which point? For this you need projective geometry, which already relies on these fundamentals.
– D S
Mar 4 at 20:28
• If you read about the history of attempted proofs of the parallel postulate, you can find this exact argument. [Of course, as noted, this argument does not prove the parallel postulate.] Mar 4 at 20:38
• If you use a smaller pencil (or a larger triangle) it is more obvious that you need to transport the pencil along the sides of the triangles. What happens to the angles when you do that? Mar 5 at 12:47

With this proposition, we begin to see what the arithmetic of magnitudes means to Euclid, in particular, how to add angles. Euclid says that the angle CBE equals the sum of the two angles CBA and ABE. So, one way a sum of angles occurs is when the two angles have a common vertex (B in this case) and a common side (BA in this case), and the angles lie on opposite sides of their common side.

In order to move the angles you rotate through to a common vertex to add them you use Book I Prop 29: the theorem that parallel lines make equal opposite angles with a transversal.

Then you are essentially using this argument from wikipedia:

The following is from
Edwin Wolfe, Non-Euclidean Geometry (1945) PDF
[the text I studied from long ago... Thanks, Dr. Underwood]

pages 40-41

The Rotation Proof
This ostensible proof, due to Bernhard Friedrich Thibaut (1775-1831) is worthy of note because it has from time to time appeared in elementary texts and has otherwise been indorsed.
...
This proof is typical of those which depend upon the idea of direction. The circumspect reader will observe that the rotations take place about different points ..., so that not only rotation, but translation, is involved.
...

Short answer: respecting the parallel postulate the internal angle sum of a triangle is $$180°$$ -- see Euclid's Elements, Book I, Proposition 32. In contrast, when disobeying it, in non-Euclidean geometries, the sum of a triangle internal angles is seldom $$180°$$.

Why? -- the parallel postulate implies coplanar lines, it does not "work" with skew lines. (For the coplanarity prerequisite there is one more reason.) Thus the "pencil test" of OP fails with triangles on a sphere for example, there it would lead to a wrong conclusion.

Additional remarks: Looking for the parallel postulate in the Internet, most sources show statements only equivalent to the fifth of Euclid's postulates, not the original wording. (German Wikipedia presents a quite funky specification: "In a plane $$\alpha$$ there is for every line $$g$$ and every point $$P$$ outside of $$g$$ ..." -- pardon, but a line and an ouside point do define a plane me think.)
It took a while until I found what is highly likely the original wording. Chiefly its implied coplanarity but also the history of the Parallelenproblem point out how the sum of a triangle internal angles $$=180°$$ is linked to the parallel postulate.

Define a 3D axis in space, at (0, 0, 0) with an X, Y and Z axis visually, and a point at (1, 1, 1) is in one of eight octant sections the axis form. Let’s trade up 1 for +# as being any positive value, and –# as any negative value. Starting with (+#, +#, +#), if we clockwise rotate all values 45 degrees, we and up at (0, 0, 0) and how? Removing the Z axis from the view and do the same rotation over again ignoring a Z value, we and rotate (+#, +#) clockwise 45 degrees and it becomes (+#, 0), Rotate another 45 degrees clockwise and it becomes (-#, -#). What significance is this? Any two axis rotations can reach any 3D point. In a 2D situation we observe the same by how one collapses no matter how many axis as it is required of rotation that two or more coordinates are involved. So, what I see is your pencil as if it really collapses. The reason is every angle has a compliment back angle, i.e. a 45 degrees angle has a 315 degrees angle clockwise or counterclockwise to reach the same conclusion, their sum being 360 degrees. Yet how can a compliment occur when there is not another angle? Like secant and cosecant are really just tangent of the first twists, viewed form the second perspective, but degrees to make things easy, sin, cos and tan are only from one perspective. While there is three one in one octant, and three 90-degree angles do not equal 180, unless one is always collapsed which is true. Viewing three axis, hide the negative x, negative y, and negatives spectrum, and you have a sort of corner in 3D made by the positive x, y and z’s. now say a point is at (+#, 0, 0) to get to any other single positive value, a point laying on the axis line (0, +#, 0) and (0, 0, +#) only two of the 3D axis are needed in rotation, and as well sometimes possibly only one but in rotation that would be incorrect to derive of, and that makes it a bit tricky, only rotating by each axis individually to make a 3D rotation is improperly left tan out of the collapsing evasion.

Types of triangles relations, as defined by a 3D corner

Now, this maybe doesn't help you to proof, but there is another way. Triangles are always collapsing and even 3D ones only have one triangle face of four theory faces total. In reality, it is a cubic face on the blackface, and the proof to that is in the volume of a triangle where the minimum number of triangles is needed to form a cube, is 12 as no prism may show two faces at any law of physics to give the cubic volume inside a measure, all prisms meet at a single point in theory.

Public Function DistanceEx(ByRef p1 As Point, ByRef p2 As Point) As Single
DistanceEx = (((p1.X - p2.X) ^ 2) + ((p1.Y - p2.Y) ^ 2) + ((p1.Z - p2.Z) ^ 2))
If DistanceEx <> 0 Then DistanceEx = DistanceEx ^ (1 / 2)
End Function

Public Function CubePerimeter(ByVal edge1 As Single, Optional ByVal edge2 As Single = 0, Optional ByVal edge3 As Single = 0) As Single
If edge2 = 0 And edge3 = 0 Then
CubePerimeter = (edge1 * 12)
Else
CubePerimeter = (edge1 * 4) + (edge2 * 4) + (edge3 * 4)
End If
End Function

Public Function CubeSurfaceArea(ByVal Edge As Single) As Single
CubeSurfaceArea = (6 * (Edge ^ 2))
End Function

Public Function CubeVolume(ByVal Edge As Single) As Single
CubeVolume = (Edge ^ 3)
End Function

Public Function TrianglePerimeter(ByRef p1 As Point, ByRef p2 As Point, ByRef p3 As Point) As Single
TrianglePerimeter = (DistanceEx(p1, p2) + DistanceEx(p2, p3) + DistanceEx(p3, p1))
End Function

Public Function TriangleSurfaceArea(ByRef p1 As Point, ByRef p2 As Point, ByRef p3 As Point) As Single
Dim l1 As Single: l1 = DistanceEx(p1, p2)
Dim l2 As Single: l2 = DistanceEx(p2, p3)
Dim l3 As Single: l3 = DistanceEx(p3, p1)
TriangleSurfaceArea = (((((((l1 + l2) - l3) + ((l2 + l3) - l1) + ((l3 + l1) - l2)) * (l1 * l2 * l3)) / (l1 + l2 + l3)) ^ (1 / 2)))
End Function

Public Function TriangleVolume(ByRef p1 As Point, ByRef p2 As Point, ByRef p3 As Point) As Single
TriangleVolume = TriangleSurfaceArea(p1, p2, p3)
TriangleVolume = ((CubeVolume((TriangleVolume ^ (1 / 3)) ^ 2)) / 12)
End Function


Pseudo code to get the volume of a 3d triangle, prism, is dependent on a 3D cube, just as a cubic measure (power of 3), is dependent on three axis. broken down to 2D a right triangle half of a square is length multiplied by height divided by two, and each half an equilateral right triangle. What really can help understand is proofing recursion in shapes. That is so easy and basic as a similar Pythagorean theorem proof also proofs recursion.

A right triangle fact, that will proof the Pythagorean theorem

Proof the Pythagorean theorem and recursion

I don't think the internal sum of a triangles angles equal 180 degrees, I think they equal 2 x PI, but with every modulus of PI actually not an angle, so -2xPI, -PI, 0, PI and 2x PI are invalid angles, just as 0 is an invalid degree. The spectrum is above zero to 360 degrees. You can express it using 0 and up to 359? To have a degree is have anything at all, so a zero is not valid. 360 no different than that, but it isn't missing a value to be an angle, all be it not moved is possible. Compared to time, another way to express angles in circles, clockwise and counterclockwise, in a 60-second, and 60-minute view like it is radian in degree form. Clocks are also not fractions, neither are radians. Fractions are another way to describe a unit whole circle, but square form, such like 5 rises over 5 run is 1, no matter the quadrant which signs the 1. Total slope is really diagonal. Yet y-intercept has us doing rise over distance in a 3D slope is more accurate, ho otherwise can it be obtained with three coordinates? Is also not FRACTIONS but technically is, more so than radian also technically is and isn't fractions. Fractions to represent angles are percentage too is not so, of a whole angle, 1 unit, or 100%, which can have a 0% and 100%, no void angles have to skip ahead for in radians to accommodate a percentage is not enough of a 90 degrees to accurately find the slope to plot a point an angle makes, it is clearly represented by a lot that is cool truly. Just individual rotations using cos and sin on each three axis will not produce the correct physics result for the space station, you have to use tan as well, and not leave the first of the three-axis view, yet recursion allows us to do so as if cheating. You really do need tan and it isn't but out there if not for APCS AB somewhere.