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Of the many nice proofs of the Pythagorean theorem, one large class is the "dissection" proofs, where the sum of the areas of the squares on the two legs is shown to be the same as the area of the square on the hypotenuse. For example:

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One generalization of the Pythagorean theorem is De Gua's Theorem, which concerns right-angled tetrhaedra:

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For such a tetrahedron, the areas $A,B,C$ of the three "legs" are related to the area $D$ of the "hypotenuse" by the formula $$ A^2 + B^2 + C^2 = D^2. $$ See here for a simple proof of this theorem. Note that both sides of this equation have units of area-squared, i.e. four-dimensional volume.

My question is:

Is it possible to prove De Gua's Theorem using a "dissection" in four dimensions?

There's no reason to be strict about the definition of a "dissection" -- any proof involving four-dimensional volume would be welcome. For example, it would certainly be interesting to have a proof of De Gua's Theorem that involves shearing in four dimensions.

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Jim Belk
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    Hilbert's 3rd Problem asked whether two polyhedra of equal volume are "equidecomposable" into the same sets of pieces. The analogous (true) result for polygons guarantees the existence of dissection proofs of the Pythagorean Theorem. I don't know the status of the 4d counterpart; of course, even if the result isn't true generally, it could hold specifically for the "squares" of faces of a right-corner tetrahedron ... for an appropriate definition of "square". (I actually toyed with this very same idea decades ago, but got nowhere.) – Blue Nov 12 '13 at 03:25
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    Incidentally, the tetrahedral result follows immediately from Heron's Formula for the area of the "hypotenuse face", wherein the square of the length of each bounding edge is replaced by the sum of squares of lengths of legs from the appropriate "leg face". – Blue Nov 12 '13 at 03:36
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    @Blue One possible definition of "square" is the Cartesian product of each right-triangle face with itself, which would be a 4-dimensional polytope. Alternatively, you could complete each right triangle into a rectangle, and then "square" this to get a 4-dimensional box whose volume is four times the square of the area of the original triangle. – Jim Belk Nov 12 '13 at 03:47
  • Hey, I learned a new word: Duoprism for the Cartesian product of two polytopes. :) – Blue Nov 12 '13 at 04:28
  • Look at this: http://i.stack.imgur.com/Wjxmq.png Then add - in addition to b+a and a+b - an additional height a+b ;) –  Nov 15 '13 at 17:21
  • @Adobe Your comment does not make sense to me. Can you elaborate? – Jim Belk Nov 15 '13 at 18:31
  • @JimBelk I'll try to show what I mean using a 3D soft. –  Nov 15 '13 at 18:41
  • To prove pythagora's theorem geometrically, 2D figures are used to view relations between squares of lengths (1D objects). I think that if we want to view geometrically relations between squares of areas we need to go to four dimensions so a drawing would not be possible. – Beni Bogosel Nov 15 '13 at 21:37
  • @BeniBogosel I agree. – Jim Belk Nov 15 '13 at 22:45
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    If I interpret your question as "is there a scissors congruence between the four dimensional polytopes $D^2$ and $A^2 \sqcup B^2 \sqcup C^2$", the answer is yes. I can take every step of the proof at cut-the-knot and replace the equality between areas by a scissors congruence and scissors congruence is an equivalence relation. (Every step is either of the form "triangle with base $b$ and height $h$ = rectangle with base $b$ and height $h/2$" or is the Pythagorean theorem, and those both have decomposition proofs.) Is that an answer to your question, or do you require something more? – David E Speyer Nov 18 '13 at 17:22
  • @DavidSpeyer This is an excellent point, and I think it roughly answers my question. Perhaps you should post this as an answer? – Jim Belk Nov 18 '13 at 17:33
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    This is the Pythagorean theorem in a dual space, and scissors congruence arguments seem like they would transpose correctly. If compatibility with congruence arguments can be shown then you don't need a new proof, just a generalization of the 2-d Pythagoras to 3-d. – zyx Nov 19 '13 at 01:35

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Reposting my comment as an answer:

If I interpret your question as "is there a scissors congruence between the four dimensional polytopes $D^2$ and $A^2 \sqcup B^2 \sqcup C^2$", the answer is yes.

I can take every step of the proof at cut-the-knot and replace the equality between volumes by a scissors congruence. Roughly speaking, every step in the argument either replaces a triangle with base $b$ and height $h$ by a $b \times (h/2)$ rectangle, or replaces the square on the hypotenuse of a right triangle by the union of the two squares on the sides. (More precisely, it does these things and then does some algebra, which can be interpreted as taking the products and disjoint unions with some other polytopes.)

Both replacing a triangle with base $b$ and height $h$ by a $b \times (h/2)$ rectangle, and replacing the square on the hypotenuse of a right triangle by the union of the two squares on the sides, have decomposition proofs. Compose all of those decompositions and you'll get a proof of this result.

David E Speyer
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    Thanks for this elegant answer! This doesn't quite deliver the insight I was fishing for (a unified proof for the Pythagorean theorem and De Gua's theorem), but it nicely answers the question that I asked, and clarifies the fact that the algebraic relationship between the two theorems is also a geometric relationship. – Jim Belk Nov 20 '13 at 18:55
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    I'll leave the question open for another day in case anyone else wants to answer. I'm not averse to awarding multiple bounties on this problem. – Jim Belk Nov 20 '13 at 18:57