Can you offer a geometrical proof of the 3D generalization of the Pythagorean Theorem?

Orient the right tetrahedron so that its edges lie along the x, y, and z axes. Let X be the vertex on the X-axis Y the y-axis vertex Z the z-axis vertex O the vertex at the origin. Also, length OX = x length OY = y length OZ = z length XY = a Let A = area OXY B = area OXZ C = area OYZ D = area XYZ Divide the tetrahedron into two with a plane that passes through the altitude of triangle OXY through O. Label the point of intersection between the altitude and line XY "P." let OP = g ZP = h now, clearly: A = ag/2 B = xy/2 C = yz/2 which means... A^2 + B^2 + C^2 = a^2g^2/4 + z^2(x^2 + y^2)/4 by the Pythagorean theorem... a^2 = x^2 + y^2 thus, the above simplifies to a^2(g^2 + z^2) however, by the Pythagorean theorem, h^2 = g^2 + z^2 thus, the above simplifies to a^2h^2/4 but D = ah/2. So, A^2 + B^2 + C^2 = D^2 this may not be completely geometrical, but at least it's elegant. I thought of this as a 3D anologue to a standard proof of the (2-D) Pythagorean theorem which uses similar triangles and the same division by altitude.

Here is a proof for any dimension greater than 1. You will decide how "geometrical" it is (picture added later): http://farm5.static.flickr.com/4022/4454649047_bd8c07af12_o.gif 2D: Let ABC is an arbitrary triangle, (n_a→), (n_b→), (n_c→) are outer normal vectors to the sides a, b, c respectively and, as usual |(n_a→)| = a, |(n_b→)| = b, |(n_c→)| = c, then (n_a→) + (n_b→) + (n_c→) = (0→) If angle C is right, then (n_a→) ⊥ (n_b→) and taking scalar squares: ((n_a→) + (n_b→))² = (-n_c→)² and using orthogonality (n_a→).(n_b→) = 0: (n_a→)² + (n_b→)² = (n_c→)² or a² + b² = c² /Pythagorean Theorem/. 3D: Let ABCD is an arbitrary tetrahedron, (n_a→), (n_b→), (n_c→), (n_d→) are outer normal vectors to faces opposite A, B, C, D respectively and, similarly |(n_a→)| = Area(BCD), |(n_b→)| = Area(CDA), |(n_c→)| = Area(DAB), |(n_d→)| = Area(ABC), then again (n_a→) + (n_b→) + (n_c→) + (n_d→) = (0→) /the sum of outer normal vectors is zero-vector even for every convex polyhedron/ If ABCD is right (faces, meeting at D are 2 by 2 perpendicular), then squaring ((n_a→) + (n_b→) + (n_c→))² = (-n_d→)² and using orthogonality we get (n_a→)² + (n_b→)² + (n_c→)² = (n_d→)² or Area²(BCD) + Area²(CDA) + Area²(DAB) = Area²(ABC) as required. For an orthogonal n-dimensional simplex x_{1} ≥ 0, x_{2} ≥ 0, . . ,x_{n} ≥ 0, x_{1}/a_{1} + x_{2}/a_{2} + . . + x_{n}/a_{n} ≤ 1, /a_{i} = const > 0/ the sum of outer normal vectors to its n-1-dimensional faces /again the length of every vector is equal to the n-1-dimensional measure ("hyper-area") of the face to which it is normal/ is zero-vector and the general relationship is obtained similarly squaring the sum of n-1 2 by 2 orthogonal vectors: the sum of the squares of n-1-dimensional measures of the 2 by 2 orthogonal faces is equal to the measure of the last face, squared.

The length of a line segment (i.e., hypotenuse) will be measured the same regardless of the orientation of the coordinate system. QED ............ I don't mean to be trite. There is a bit of perspiration in this bold statement. The definition of a circle is 1 = (x/r)^2 + (y/r)^2. For a ball you add a z-coordinate, and you keep adding coordinates as you. [I also realize that the the definition of a circle is dependent on the pythagorean theorem -- but perhaps there is something deeper in the choice of perpendicular coordinates] Once you have a line segment, pick one end as the center of the circle/ball/4d-ball. You can then rotate the coordinate system however you like and you will always end up with: r^2 = x^2 + y^2 + z^2 ...