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 $XYZT$ be our tetrahedron, where $TX\perp TY$, $TX\perp TZ$, $TY\perp TZ$, $TX=x$, $TY=y$ and $TZ=z$.
Thus, $XY=\sqrt{x^2+y^2}$, $XZ=\sqrt{x^2+z^2},$ $YZ=\sqrt{y^2+z^2}$ and
$$S_{\Delta XYZ}=\frac{1}{4}\sqrt{\sum_{cyc}(2(x^2+y^2)(x^2+z^2)-(x^2+y^2)^2)}=$$
$$=\frac{1}{4}\sqrt{\sum_{cyc}(2x^4+6x^2y^2-2x^4-2x^2y^2)}=\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}$$ and since
$$\left(\frac{1}{2}xy\right)^2+\left(\frac{1}{2}xz\right)^2+\left(\frac{1}{2}yz\right)^2=\left(\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}\right)^2,$$ we are done!
