Find elegant proof of 3D generalization of Pythagorean theorem?

Let (cosα,cosβ,cosγ) be direction cosines of a plane that slices off a solid like this in O1 and meets axes at X,Y,Z. Let area of triangle XYZ be Δ.

cos²α+cos²β+cos²γ = 1 → Δ² = (Δcosα)² + (Δcosβ)² + (Δcosγ)²

But angle between yz plane and XYZ is α and so projection of Δ onto yz plane is Δcosα.

∴ area OYZ = Δcosα with similar results for other faces.

Without loss of generality, let the four vertices of the given tetrahedron be, D (0, 0, 0), A(a, 0, 0), B(0, b,0) and C(0, 0, c)

We have A = (1/2)bc, B = (1/2)ac an C= (1/2)ab and it can be proved that

D = (1/2)√(a²b² + b²c² c²a²). So it follows that

D² = A² + B² + C²

_____________________________________________

Above A, B and C meant areas and also points A, B C and D

Proof of area D of the triangle ABC is as follows, a, b c are line segments, and A, B and C are angles.

The triangle ABC has three sides of lengths:

√(a² + b²), √(b² + c²) and √(c² + a²)

D = Area of triangle ABC = (1/2)*{√(b² + c²)}*{√(c² + a²)}* sin C ----------- (1)

Now Cos C = [{b² + c² + c² + a² -(a² + b²)}/{2*{√(b² + c²)}*{√(c² + a²)}}]

so sin C = √(1 - cos² C) = [{√(a²b² + b²c² c²a²)}/{√(b² + c²)}*{√(c² + a²)} ----(2) from (1) and (2), we do get

D = (1/2)*√(a²b² + b²c² c²a²)}/

imagine a rectangular block with sides A, B, C and find the side D that is the straight line between two diagonally apposed corners.

First you look at the face made by A and B and compute the diagonal E by E^2 = A^2 + B^2

Then you notice that the line E also makes a right angle with C so that D^2 = C^2 + E^2

Therefore D^2 = C^2 + A^2 + B^2

May I offer a very belated proof, relying on "known facts", on a question CLOSED3 weeks ago? Oddly enough Y!A will let me do it NOW!

Let a→, b→, c→, d→ are outer normal vectors to the faces of the tetrahedron;

let as usual |a→| = A, |b→| = B, |c→| = C, |d→| = D; then

Known fact #1: (w→)² = |w→|² for every vector w→;

Known fact #2: a→ b→ = b→c→ = c→a→ = 0 (orthogonality);

Known fact #3: a→ + b→ + c→ + d→ = 0→.

Squaring both sides of a→ + b→ + c→ = -d→ we get what we need and similarly we can generalize the result for higher dimensions.

Without loss of generality, let the vertices be located at (0,0,0), (a,0,0), (0,b,0), (0,0,c), with a, b, and c positive.

Let A = area of right triangle with vertices (0,0,0), (0,b,0), (0,0,c).

Let B = area of right triangle with vertices (0,0,0), (a,0,0), (0,0,c).

Let C = area of right triangle with vertices (0,0,0), (a,0,0), (0,b,0).

Let D = area of triangle with vertices (a,0,0), (0,b,0), (0,0,c).

Clearly A = bc/2, B = ac/2, and C = ab/2.

To find D, we can use the fact that the area of a triangle, with two of its sides formed by vectors u and v, is half the magnitude of the cross product of u and v.

Two of the sides of the triangle with vertices (a,0,0), (0,b,0), (0,0,c) are formed by the vectors u = <-a,b,0> and v = <-a,0,c>. Therefore,

D = (1/2)|u cross v|

= (1/2)|<-a,b,0> cross <-a,0,c>|

= (1/2)|<det[b 0; 0 c], -det[-a 0; -a c], det[-a b; -a 0]>|

= (1/2)|<bc, ac, ab>|

= (1/2)sqrt((bc)^2 + (ac)^2 + (ab)^2).

Therefore,

D^2 = (1/2)^2 ((bc)^2 + (ac)^2 + (ab)^2)

= (bc/2)^2 + (ac/2)^2 + (ab/2)^2

= A^2 + B^2 + C^2.

(Note: the first answerer misinterpreted A, B, C, D as lengths instead of areas.)

May Jesus richly bless you today and Happy New Year!