(Previously part of

The Triangle Sum Theorem

Old and new proofs concerning the sum of interior angles of a triangle.
(More on the hidden depths of triangle qualia.)

Aaron Sloman
Installed: 9 Sep 2012
Please report bugs to:
This was originally part of the file:
That file is mainly about areas, so this portion, concerned with angles, was moved here on 28th May 2013.

Last updated:
25 Apr 2016; 8 Sep 2017; 23 Sep 2017;
26 Feb 2015: added link to document showing how in P-geometry an arbitrary angle
can be trisected.
29 May 2013 ....Updates as part of the original file deleted...


This file is
PDF derivative:

Related documents
A partial index of discussion notes in this directory is in

See also this discussion of "Toddler Theorems":

This document illustrates some points made in a draft, incomplete, discussion of transitions in information-processing, in biological evolution, development, learning, etc. here. That document and this one are both parts of the Meta-Morphogenesis project, partly inspired by Turing's 1952 paper on morphogenesis.

James Gibson's theory of perception of affordances, is very closely related to mathematical perception of structures, possibilities for change, and constraints on changes (structural invariants). Gibson's ideas are summarised, criticised and extended here:

This discussion of theorems about processes that alter or preserve areas of triangles is closely related:

These discussions draw attention to common confusions about the nature of embodied
cognition in 'enactivist' theories, and illustrate the need to distinguish 'online intelligence'
from 'offline intelligence'.

Related Video On Adam Ford's Web Site
At the AGI conference in Oxford, December 2012, Adam Ford interviewed me about this and related topics. I used the triangle sum theorem as an example in the interview, available at

Mary Pardoe's Twitter page:
Some of Vi Hart's wonderful mathematical video doodles are also relevant:


---------- robot geometer
When will the first baby robot grow up to be a mathematician?



The Triangle Sum Theorem

The triangle sum theorem is normally expressed as "The interior angles of a triangle add
up to 180 degrees". This assumes a standard way of measuring angles, according to which a
complete rotation would be 360 degrees and a half rotation 180 degrees. But we can
equivalently express the theorem as "The interior angles of a triangle add up to a
straight line", or "... half a rotation", which does not require any conventional unit for
measuring angles". As we'll see below that suggests a way of proving the theorem by
considering a succession of rotations and seeing what they add up to, an idea suggested
many years ago by Mary Pardoe after she had become a mathematics teacher.

There is a standard way (or small set of standard ways) of proving the theorem

Triangle Sum Theorem (TST): The interior angles of a triangle add up to a
straight line, or half a rotation (180 degrees).

These standard methods all make use of some version of Euclid's parallel postulate,
(Axiom 5 in Euclid's elements) which can be formulated in several equivalent ways, e.g.

Two straight lines L1 and L2 are parallel if and only if they are co-planar and
have no point in common, no matter how far they are extended.

Given a straight line L in a plane, and a point P in the plane not on L, there
is exactly one line through P that is in the plane and parallel to L.
(That was not Euclid's formulation, but is perhaps intuitively the clearest formulation.)

All of this presupposes the concept of "straightness" of a line. For now I'll take
that concept for granted, without attempting to define it, though we can note that if
a line is straight it is also symmetric about itself (it coincides with its
reflection) and also it can be slid along itself without any gaps appearing. If it
were possible to view a straight line from one end it would appear as a point.

The "standard" ways of proving the TST make use of properties of angles formed
when a straight line joins or crosses a pair of parallel lines:

COR: Corresponding angles are equal:
If two lines L1, L2 are parallel and a third line L3 is drawn from any point P1
on L1 to a point P2 on L2 and continued beyond P2,
then the angle that L1 makes with the line L3 at point P1, and the angle L2
makes with the line L3 at point P2 (where the angles are on the same side of
both lines) are equal.

ALT: Alternate angles are equal:
If two lines L1, L2 are parallel and a third line L3 is drawn from any point P1
on L1 to a point P2 on L2,
then the angle L1 makes with the line L3 at point P1, and the angle L2 makes
with the line L3 at point P2 (on the opposite sides of both lines) are equal.

For more on transversals and relations between the angles they create, see
That page teaches concepts with some interactive illustrations, but presents no proofs.

The Euclidean proofs of COR and ALT are presented here:


The "standard" proofs of the "Triangle Sum Theorem"

Two "standard" proofs of the triangle sum theorem using parallel lines, and the
Euclidean theorems COR and/or ALT stated above, are shown below in Figure Ang1:

Figure Ang1: two proofs

Warning: I have found some online proofs of theorems in Euclidean geometry with bugs
apparently due to carelessness, so it is important to check every such proof found
online. The fact that individual thinkers can check such a proof is in part of what
needs to be explained.


Mary Pardoe's proof of the Triangle Sum Theorem

Many years ago at Sussex university I was visited by a former student Mary Pardoe, who had been teaching mathematics in schools. She told me that her pupils had found the standard proof of the triangle sum theorem hard to take in and remember, but that she had found an alternative proof, which was more memorable, and easier for her pupils to understand.
Note: In the original publication reporting this proof I mistakenly referred to the author as Mary Ensor, her name as a student. I think she was already Mary Pardoe at the time she visited me.

Her proof just involves rotating a single directed line segment (or arrow, or pencil, or ...) through each of the angles in turn at the corners of the triangle, which must result in its ending up in its initial location pointing in the opposite direction, without ever crossing over itself.

So the total rotation angle is equivalent to a straight line, or half rotation, i.e. 180 degrees, using the convention that a full rotation is 360 degrees.

The proof is illustrated below in Figure Ang2.

Figure Ang2: rotating segment

In order to understand the proof, think of the blue arrow, labelled "1", as starting on line AC, pointing from A to C, and then being rotated first around point A, then point B, then point C until it ends up on the original line but pointing in the direction of the dark grey arrow, labelled "4".

So, understanding the proof involves considering what happens if

A "time-lapse" presentation of the proof may be clearer, as shown in Figure Ang3:

Figure Ang3:
--- rotating segment

It may be best to think of the proof not as a static diagram but as a process, with stages represented from left to right in Figure Ang3. In the first stage, the pale blue arrow starts on the bottom side of the triangle, pointing to the right then is rotated through each of the internal angles A, B, C, always rotated in the same direction (counter-clockwise in this case), so that it lies on each of the other sides in succession, until it is finally rotated through the third angle, c, after which it lies on the original side of the triangle, but obviously pointing in the opposite direction. Some people may prefer to rotate something like a pencil rather than imagining a rotation depicted by snapshots.

In this triangle the sides are not very different in length, which conceals a problem that can arise if the first side the arrow is on is very short and the other two are longer. If the length of the arrow is fixed by the length of the first side, you would need to imagine either that the arrow stretches or shrinks as it rotates, or that it slides along a line after reaching it so as to be able to rotate around the next vertex. Alternatively you can imagine that the depicted arrow is part of a much longer invisible arrow, so that, as the invisible arrow rotates from one side to another, it always extends beyond both ends of the new side, and can then rotate around the next vertex. I leave it to the reader to think about these alternatives and what difference they make to the proof, and to the cognitive competences required to construct and understand the proof.

For an arrow to be rotated in a plane and end up lying in its original position it must have been rotated through some number of half-rotations. (Each half rotation brings it back to the original orientation, but pointing in alternate directions.)

Since (1) the arrow at no point crossed over its original orientation, and (2) it ended up pointing in the opposite direction to its original orientation, the total rotation was through a half circle -- which is clear if you actually perform the rotations using a physical object, such as a pencil.

And since that rotation was made up of combined rotations through angles A, B, and C, those three angles must add up to a half circle, i.e. 180 degrees.

A crucial feature of our ability to think about the diagram and the process, is that we (presumably including you, the reader) can see that the key features of the process could have been replicated, no matter what the size or orientation of the triangle, no matter what the lengths of the sides or the sizes of the angles, no matter which side the arrow starts on, no matter which way it is pointing initially, and no matter in which order the rotations are performed, e.g. A then B then C, or C reversed, then B reversed, then A reversed.

This proof of the triangle sum theorem, using a rotating moving arrow, works for all possible triangles on a plane -- as do the standard Euclidean proofs using parallel lines.

(Works only in some web browsers.)
rotating segment

On a plane surface, rotating the blue arrow through the three internal angles
(i.e. A, then B, then C) always brings it back to the starting line, pointing in
the reverse direction, without ever crossing over its original orientation,
and this (obviously?) doesn't depend on the shape of the triangle.

This proof is unlike standard proofs in Euclidean geometry since it involves consideration of continuous processes, and therefore involves time and temporal ordering, whereas Euclidean geometry does not explicitly mention time or processes -- though there are some theorems about the locus of point or line satisfying certain constraints, which can be interpreted either as specifying properties of processes extended in time, or as properties of static trajectories, e.g. properties of lines or curves.

presents a more detailed, but still incomplete, discussion, of the geometrical prerequisites for some of the above reasoning. It introduces the idea of P-geometry, which is intended to be Euclidean geometry without the Axiom of Parallels (Euclid's Axiom 5), but with time and motion added, including translation and rotation of rigid line-segments.
NOTE (Added 8 Sep 2017):
Mary Pardoe remains actively involved in mathematics education.
Her twitter site is a steady stream of information:


Is the Pardoe proof valid?

NOTE: I have presented Mary Pardoe's proof in several places, over several years, e.g.
Aaron Sloman, 2008,
Kantian Philosophy of Mathematics and Young Robots, in Intelligent Computer Mathematics, Eds. Autexier, S., Campbell, J., Rubio, J., Sorge, V., Suzuki, M., and Wiedijk, F., LLNCS no 5144, pp. 558-573, Springer,
(This paper referred to Mary Ensor.)

Aaron Sloman, 2010,
If learning maths requires a teacher, where did the first teachers come from?, In Proceedings Symposium on Mathematical Practice and Cognition, AISB 2010 Convention, De Montfort University, Leicester

And in talks on mathematical cognition and philosophy of mathematics here:

The presentations produced no responses -- either critical or approving, except that in one informal discussion a mathematician objected that the proof was unacceptable because the surface of a sphere would provide a counter example. However, the surface of a sphere provides no more and no less of a problem for Pardoe's proof than for the standard Euclidean proofs since both proofs are restricted to planar surfaces.

I tried searching for online proofs to see if anyone else had discovered this proof or used it, but nothing turned up. The proof using rotation is so simple and so effective that both Mary Pardoe and I feel sure it must have been discovered previously.

NOTE ADDED 6 Oct 2012 (Asperti and Scott):
I have discovered that as a result of the discussion in 2010 on the MKM-IG email list, Andrea Asperti mentioned the proof (and the email discussion) in this paper, discussing related issues:

Andrea Asperti, Proof, Message and Certificate,
in AISC/MKM/Calculemus, 2012, pp. 17--31,
And in this slide presentation with the same title, starting with Mary's proof, and a comment on the proof by Dana Scott:
Dana Scott commented:
"The proof is fine and really is the same as the classical proof. To see this, translate (by parallel translation) all the three angles of the triangle up to the line through the top vertex of the triangle parallel to the lower side."
He later added
"I should have commented in my explanation of the proof that if you translate the line on which the base of the triangle sits along each of the sides up to the vertex, then both actions result in the same line - the unique parallel."
Is it really the same as the classical proof?
But this still leaves several questions open: Why did neither Mary, nor her students, nor others to whom I have shown the proof not claim that they could accept it ONLY by relating it to a proof that depends explicitly on Euclid's parallel axiom; and why should anyone regard as inferior the direct intuition provided by this proof that the three internal angles add up to half a rotation?

I suspect we shall not have good answers to these questions until we have a much deeper understanding of the combination of biological geometric reasoning mechanisms produced by evolution plus the (epigenetic) processes of individual development leading up to use of those mechanisms -- deep enough to build a baby robot that can grow up to have the competences of ancient mathematicians.

We must not forget that however those competences are eventually explained they were of tremendous importance for human beings, not least because the contents of Euclid's Elements are still in use by engineers, scientists and mathematicians all around the planet, every day. Moreover, the kinds of learning involved seem to be beyond the scope of current (e.g. 2017) AI robot designs and learning mechanisms, e.g. "deep learning" that depends on probabilistic reasoning. (This topic is discussed further in

The P-geometry document begins to specify a variant of Euclidean geometry without the parallel axiom, but allowing for translation and rotation of line segments while maintaining their length.

P-geometry (not yet fully specified) is used to trisect an arbitrary angle.

There is a "process" version of the proof of Pythagoras theorem that makes use of a video. A version implemented in Pop-11 is illustrated in the video in this tutorial:
The video attempts to demonstrate the invariance by showing how the shapes and or sizes of the triangles, squares and rectangles can be changed without changing the structural relationships. This was inspired by a demonstration originally provided by Norman Foo, using different transformations:
One of the striking facts about Pythagoras' theorem is how many different ways it can be, and has been, proved.

NB: The programs that present such proofs do not themselves understand the proofs. They can be powerful "cognitive prosthetics" for humans learning mathematics, but the programs do not know what they have done, or why they have done it, and do not understand the invariants involved -- e.g. essentially the same proof could have started with a triangle with different angles, or a triangle of a different size.


Added 9 Feb 2013: Another proof of the sum theorem, by Kay Hughes
Modified 4 Mar 2013:

In February 2013, at an orienteering club dinner, I was talking about education with Kay Hughes and asked if she could remember how to prove the Triangle Sum Theorem. She could not remember a proof, but quickly thought up a proof that I had never previously encountered. My presentation here does not use her words, but offers a more explicit elaboration of the ideas she presented. The key idea was to use a theorem about the sum of external angles of a polygon always being a whole rotation (360 degrees) and combining that with the fact that each of the external angles has an internal angle as complement, as explained in more detail below.
Figure Ang4: Proof by Kay Hughes

It should be obvious from the figure that it presents a proof that the exterior anti-clockwise angles of a triangle (A+B+C) sum to a circle (360 degrees) as do the exterior clockwise angles, not shown in the figure.
Added 19 Mar 2013: This was named "The total turtle trip theorem" by Seymour Papert, in his Mindstorms: Children, Computers, and Powerful Ideas (1978), though it was well known long before then. (It can be generalised to smooth simple closed curves. See also .)

The exterior anti-clockwise angles are those obtained by extending each side in turn in one direction then rotating the extension to line up with the next side. So, for example, in Figure Ang4, the internal angles are a, b and c; whereas the exterior anti-clockwise angles A, B and C are got by extending the first side to location 1 then rotating the extension through angle A to the next side, then extending that side to location 2 and rotating the extension through angle B to the second side, and so on.

Because results of all those rotations bring the rotated arrows back to the original orientation, indicated at 1 in the figure, and the rotated arrow does not pass through its original direction, the total external anti-clockwise rotation must be a full circle (i.e. 360 degrees). An exercise left to the reader is to show that that's true not only for triangles but for all polygons, and, by symmetry, must also be true for the sum of the clockwise external angles. So:
Theorem External: A + B + C = 360

But each of the internal angles is the complement of the adjacent internal angle, because they sum to a straight line. So we have these three truths:
Theorem: A + a = 180 therefore a = 180 - A
Theorem: B + b = 180 therefore b = 180 - A
Theorem: C + c = 180 therefore c = 180 - A
So, the sum of the internal angles is
   a + b + c = (180 - A) + (180 - B) + (180 - C)
= 180 + (180 + 180) - (A + B + C)
= 180 + 360 - (A + B + C)
Then substituting from Theorem External:
= 180 + 360 - 360
= 180
So, we have another proof of the standard Triangle Sum Theorem:
Theorem Internal: a + b + c = 180

I tried searching for that proof using google and did not find a previous occurrence of it, though there seem to be many web sites that mention both the triangle sum theorem for interior angles and the theorem about exterior angles always summing to 360.

NOTE (Added 25 Apr 2016): Michael Fourman
informs me that he encountered the external angle proof while at school.



See the acknowledgements section of the paper on P-Geometry

Offers of help in making progress will be accepted gratefully, especially suggestions regarding mechanisms that could enable robots to have an intuitive understanding of space and time that would enable some of them to rediscover Euclidean geometry, including Mary Pardoe's proof.

I believe that could turn out to be a deep vindication of Immanuel Kant's philosophy of mathematics. Some initial thoughts are in my online talks, including
Why (and how) did biological evolution produce mathematicians?
Video presentation with online notes:
Why can't (current) machines reason like Euclid or even human toddlers?
(And many other intelligent animals)
Prepared for AGA Workshop at IJCAI 2017.


Maintained by
Aaron Sloman
School of Computer Science
The University of Birmingham