1. ## Three-dimensional Pythagoras' theorem

Suppose you have a right-angled tetrahedron, i.e. a tetrahedron with a vertex where three faces meet at right angles. Let $A$ be the area of the face opposite this vertex, and $B, C, D$ the areas of the remaining faces. Show that $A^2=B^2+C^2+D^2$.

Edit : I had first posted a more general problem but I'll put it on hold for the time being.

2. Spoiler:
Let

sides of $A:x,y,z$
sides of $B: \alpha, \gamma, y$
sides of $C: \alpha, \beta, x$
sides of $D: \beta, \gamma, z$.

Then we have

$B=\frac{1}{2}\alpha \gamma, \, C=\frac{1}{2}\alpha \beta , \, D=\frac{1}{2}\beta \gamma$

and

$x^2=\alpha^2+\beta^2$
$y^2=\alpha^2+\gamma^2$
$z^2=\beta^2+\gamma^2.$

Using Heron's formula, we have

$A^2=\frac{1}{16}(x+y+z)(y+z-x)(x+y-z)(z+x-y)$

$=\frac{1}{16}((y+z)+x)((y+z)-x)(x+(y-z))(x-(y-z))$

$=\frac{1}{16}((y+z)^2-x^2)(x^2-(y-z)^2)$

$=\frac{1}{16}(2yz-(x^2-y^2-z^2))(2yz+(x^2-y^2-z^2))$

$=\frac{1}{16}(4y^2z^2-(x^2-y^2-z^2)^2)$

$=\frac{1}{16}\left[4(\alpha^2+\gamma^2)(\beta^2+\gamma^2)-((\alpha^2+\beta^2)-(\alpha^2+\gamma^2)-(\beta^2+\gamma^2))^2\right]$

$=\frac{1}{16}\left[4(\alpha^2\beta^2+\beta^2\gamma^2+\alpha^2\gamma^2 +\gamma^4)-4\gamma^4\right]$

$=\frac{1}{4}\alpha^2\beta^2+\frac{1}{4}\beta^2\gam ma^2+\frac{1}{4}\alpha^2\gamma^2=C^2+D^2+B^2.$

3. Here's my solution

Spoiler:
Couldn't you just put your tetrahedron in the $xyz$ planes where the $xy, yz \;\text{and}\; xz$ plane are three of the four sides. Let the face A be described by the equation $ax + by + cz = 1$. If so then the area of the sides B, C and D are $\frac{1}{2} \frac{1}{ab}, \frac{1}{2} \frac{1}{ac} \; \text{and}\; \frac{1}{2} \frac{1}{bc}$. So

$
B^2+C^2+D^2 = \frac{1}{4} \left( \frac{1}{a^2b^2} + \frac{1}{a^2c^2} + \frac{1}{b^2c^2} \right)
$

The area of face $A$ is $\iint \limits_R \sqrt{1 + z_x^2 + z_y^2}dA = \sqrt{1 + \frac{a^2}{c^2} + \frac{b^2}{c^2}} \frac{1}{2ab}$

and so $A^2 = \left(1 + \frac{a^2}{c^2} + \frac{b^2}{c^2}\right) \frac{1}{4a^2b^2} = B^2 + C^2 + D^2$

4. Let

$A = A(a,0,0) , B = B(0,b,0) , C = C(0,0,c)$

together with the origin , they form a right-angled tetrahedron

we have $CA = (a,0,-c) , CB = (0,b,-c)$

and their cross product is

$CA \times CB = (a,0,-c) \times (0,b,-c)$

$= ( bc , ac , ab )$

we also know that the area of the face opposite this vertex $S$ is

the half of the modulus of the cross product so

$S^2 = \frac{(ab)^2 + (bc)^2 + (ac)^2 }{4}$

$= S^2_1 + S^2_2 + S^2_3$

where $S_1 , S_2 , S_3$ are the area of the lateral faces of the tetrahedron .