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

Given a right tetrahedron, where the right angles of 3 right triangle faces share a common vertex, let A, B, C be the areas of those 3 right triangle faces, and D be the area of the 4th face. Then the 3D generalization of the Pythagorean theorem is: A² + B² + C² = D² This can be proven algebraically without... show more Given a right tetrahedron, where the right angles of 3 right triangle faces share a common vertex, let A, B, C be the areas of those 3 right triangle faces, and D be the area of the 4th face. Then the 3D generalization of the Pythagorean theorem is:

A² + B² + C² = D²

This can be proven algebraically without too much trouble (however tedious). But the famous Pythagorean theorem has many geometrical proofs. Can you devise a geometrical proof of the 3D generalization of the Pythagorean theorem?

10 points goes to the "most geometrical" proof that avoids algebraic work as much as possible. Good luck!
Update: An abstract proof in 4D space is okay.
Update 2: Good try, Ben. I will leave this open to give others a chance to give even more elegant geometrical proofs along the lines of the original classical Greek proof of it, which has no algebraic manipulation.
Update 3: Such as several posted in Woflram's site about this:

http://mathworld.wolfram.com/Pythagorean...
Update 4: ginalino, nice proof, even though that MH ⊥ JKL is not immediately apparent.
Update 5: AI P, I had a couple brute force algebraic ways of proving this before devising this Y!A question. Yours is one of them. It's pretty straightforward, although not as pretty as I had hoped.
Update 6: Duke, for a lot of reasons, I especially liked your answer. I will comment later after others have posted their versions.
Update 7: FGR, it's the sum of squares of AREAS. THAT 3D version.
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