De Gua's Theorem

De Gua's theorem says that if three edges of a tetrahedron meet at right angles at one corner, like the corner of a box, then the square of the area of the face opposite that corner equals the sum of the squares of the areas of the three right-angled faces: D² = A² + B² + C². It is the Pythagorean theorem for areas, one dimension up.

The theorem is named after the French mathematician Jean Paul de Gua de Malves, who published it in 1783. Below you will find the statement, three different proofs, worked examples you can check by hand, and how the result fits into a family of "Pythagorean theorems" in every dimension.

The statement

A right-corner tetrahedron (also called a trirectangular tetrahedron) has a vertex O where three edges OP, OQ and OR are mutually perpendicular. Call their lengths p, q and r. Three of its faces are right triangles that meet at O; the fourth face, PQR, is the slanted face opposite the corner.

Let A, B and C be the areas of the right faces OPQ, OQR and ORP, and D the area of the slanted face PQR. Because each right face is a right triangle, A = 1 over 2pq, B = 1 over 2qr and C = 1 over 2rp. De Gua's theorem states:

The analogy with the ordinary theorem is exact, one dimension up:

Do not confuse this with the other "3D Pythagorean theorem", d² = p² + q² + r² for the space diagonal of a box (see the Pythagorean theorem in 3D). That one is about lengths; de Gua's is about areas.

Proof 1: with the cross product

Put O at the origin and the three edges along the axes: P = (p, 0, 0), Q = (0, q, 0), R = (0, 0, r). The area of a triangle is half the length of the cross product of two of its edge vectors. For the slanted face, use PQ = (−p, q, 0) and PR = (−p, 0, r):

So D = 1 over 2√q²r² + p²r² + p²q², and squaring,

Look at the three components of (qr, pr, pq): they are exactly twice the areas of the right faces. That is no accident. The components of a face's area vector are the areas of its shadows on the three coordinate planes, and here those shadows are the right faces — which is what the animation shows. The theorem is then just the three-dimensional length formula applied to the area vector.

Proof 2: with the ordinary Pythagorean theorem only

This proof needs nothing beyond right triangles, so it is within reach of a high-school geometry course.

1. Drop an altitude in the back face. In the right triangle OQR, let H be the foot of the perpendicular from O to QR. Its height is OH = qr over √q² + r² (twice the area divided by the base), and QR = √q² + r².
2. Show PH is the altitude of the slanted face. The edge OP is perpendicular to the whole plane OQR, so QR ⊥ OP. Also QR ⊥ OH. A line perpendicular to two crossing lines of a plane is perpendicular to the plane, so QR is perpendicular to the plane POH, and in particular QR ⊥ PH.
3. Use Pythagoras in triangle POH. It has a right angle at O, so PH² = p² + OH² = p² + q²r² over q² + r².
4. Compute the area. D² = 1 over 4 QR² · PH² = 1 over 4(q² + r²)(p² + q²r² over q² + r²) = 1 over 4(p²q² + p²r² + q²r²), which is A² + C² + B².

A third route is brute force: compute the three sides of PQR with the ordinary theorem, √p² + q², √q² + r² and √r² + p², and feed them into Heron's formula. The algebra is longer but it lands on the same expression.

Worked examples

More facts about the right-corner tetrahedron

• The slanted face is always acute. Its sides satisfy PQ² + QR² − RP² = 2q² > 0, and similarly at the other corners, so by the side test for acute triangles all three of its angles are less than 90°. In Example 1 they are about 85.6°, 55.3° and 39.1°.
• A reciprocal Pythagorean theorem for the height. If h is the distance from O to the slanted face, then 1 over h² = 1 over p² + 1 over q² + 1 over r². For Example 1, 1 over 4 + 1 over 36 + 1 over 64 = 169 over 576, so h = 24 over 13 ≈ 1.846. This matches the volume: V = 1 over 6pqr = 16 and also V = 1 over 3Dh = 1 over 3(26)(24 over 13) = 16.
• A law of cosines for any tetrahedron. If the corner is not a right angle, a correction term appears, just as the law of cosines corrects the Pythagorean theorem: D² = A² + B² + C² − 2(ABcosθ_AB + BCcosθ_BC + CAcosθ_CA), where θ_AB is the dihedral angle between faces A and B, and so on. With three right dihedral angles at the corner, every cosine is 0 and de Gua's theorem remains.

Higher dimensions

The pattern continues. In n dimensions, a simplex (the analog of a triangle or tetrahedron) with a corner where all n edges are mutually perpendicular satisfies: the squared (n − 1)-dimensional volume of the face opposite the corner equals the sum of the squared volumes of the other n faces. The ordinary theorem is the case n = 2 and de Gua's theorem is n = 3. This general version was proved by P. S. Donchian and H. S. M. Coxeter in 1935.

An even broader statement was given by Donald R. Conant and William A. Beyer in 1974: for any flat k-dimensional region in n-dimensional space, its squared k-volume equals the sum of the squared k-volumes of its projections onto all the k-dimensional coordinate subspaces. The "shadow" picture in the animation above is the case k = 2, n = 3. See the Pythagorean theorem in n dimensions for the length version, and the theorem for other shapes for a different way to generalize the squares.

Who was Jean Paul de Gua de Malves?

Jean Paul de Gua de Malves (1713–1785; sources differ by a year on both dates) was a French priest — hence the title abbé — and mathematician from the Languedoc region. His father was ruined by speculation in John Law's ill-fated Mississippi Company, so the family fortune was gone before his career began.

• In 1740 he published Usages de l'analyse de Descartes, a study of tangents, asymptotes and singular points of algebraic curves using algebra rather than calculus.
• He gave a proof of Descartes' rule of signs, which counts the possible positive roots of a polynomial.
• He was elected a Fellow of the Royal Society of London in 1743 and was a member of the Paris Academy of Sciences.
• He was an early editor of the French Encyclopédie, before Denis Diderot took over the project in 1747.

He published the tetrahedron theorem in 1783, but he was not the first to know it. It had been found long before by Johann Faulhaber (1580–1635) and René Descartes (1596–1650), and Charles de Tinseau d'Amondans published a slightly more general version at about the same time as de Gua. As often happens, the name stuck to the person whose publication became best known. For the much older story of the two-dimensional theorem, see the history of the Pythagorean theorem.