Show that $A^2 +B^2+C^2=D^2$ using the following diagram (tetrahedron) Slicing a corner off a square gives a right-angled triangle, as shown in the diagram below.
The lengths of the sides of this triangle are related by Pythagoras’s theorem: $a^
2 + b^
2 = c^
2$
. Show
that this two-dimensional setup generalises to three dimensions in the following way. Slice a corner
off a cube, as shown in the diagram below. This gives a tetrahedron in which three of the faces are
right-angled triangles, while the fourth is not. Let’s call the areas of the three right-angled faces $A,
B, C$ and the area of the fourth face $D$. 
$A
^2 + B^
2 + C^
2 = D^2$
.
Can anyone explain to me what formula/how to go about doing this question? thank you.

Let the sliced corner be at the origin and the fourth face lie in the plane $x/a + y/b + z/c = 1$. The foot of the altitude drawn to the fourth face is at $\lambda (1/a, 1/b, 1/c), \,\lambda = 1/(1/a^2 + 1/b^2 + 1/c^2)$. Therefore
$$9 V^2 = A^2 a^2 = B^2 b^2 = C^2 c^2 = D^2 h^2, \\
A^2 + B^2 + C^2 =
D^2 h^2 \left( \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \right) =
D^2.$$
