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.
 A: 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!
A: Let $OXYZ$ be the tetrahedron where $a:=OX$, $b:=OY$, and $c:=OZ$ are orthogonal vectors in $\mathbb{R}^3$ with respect to its usual inner product.  Then, note that
$$A:=\frac{1}{2}\,\|b\times c\|\,,\,\,B:=\frac{1}{2}\,\|c\times a\|\,,\text{ and }C:=\frac{1}{2}\,\|a\times b\|$$
are the areas of the triangle $OYZ$, $OZX$, and $OXY$, respectively.  Here, $\times$ is the usual cross product and $\|\_\|$ is the Euclidean norm (induced by the standard inner product) of $\mathbb{R}^3$.  Prove that the area of the triangle $XYZ$ is
$$D:=\frac{1}{2}\,\big\|(b-a)\times (c-a)\big\|=\frac{1}{2}\,\|b\times c+c\times a+a\times b\|\,.$$
Finally, prove that
$$b\times c\,,\,\,c\times a\,,\text{ and }a\times b$$
are mutually orthogonal (i.e., $a\parallel b\times c$, $b\parallel c\times a$, and $c\parallel a\times b$).  It follows immediately that
$$A^2+B^2+C^2=D^2\,,$$
since $\|p+q+r\|=\sqrt{p^2+q^2+r^2}$ for mutually orthogonal vectors $p,q,r\in\mathbb{R}^3$.

In general, let $n>1$ be an integer and consider the $n$-simplex $OX_1X_2\ldots X_n$ in $\mathbb{R}^{n}$, where $a_i:=OX_i$ for $i=1,2,\ldots,n$ are mutually orthogonal vectors in $\mathbb{R}^n$ with respect to its standard inner product.  Suppose that $e_1,e_2,\ldots,e_n$ are standard basis vectors of $\mathbb{R}^n$.  Identify the exterior power
$\bigwedge^{n-1}\mathbb{R}^n$ as $\mathbb{R}^n$ via the identification
$$e_1\wedge e_2 \wedge\ldots \wedge e_{i-1} \wedge e_{i+1} \wedge \ldots \wedge e_n = (-1)^{i+1}e_i\,.$$
(See this link for more detail.)  This identification induces an isometric isomorphism $\bigwedge^{n-1}\mathbb{R}^n\cong\mathbb{R}^n$.  From now on, we just say that $\bigwedge^{n-1}\mathbb{R}^n=\mathbb{R}^n$, via the identification above.  Hence, $\bigwedge^{n-1}\mathbb{R}^n$ inherits the Euclidean norm $\|\_\|$ from $\mathbb{R}^n$.
For $i=1,2,\ldots,n$, the $(n-1)$-volume of the $(n-1)$-simplex $$S_i:=OX_1X_2\ldots X_{i-1}X_{i+1}\ldots X_n$$ is given by
$$v_i:=\frac{1}{(n-1)!}\,\left\| A_i\right\|\,,$$
where 
$$A_i:=a_1\wedge a_2\wedge \ldots \wedge a_{i-1} \wedge a_{i+1} \wedge \ldots \wedge a_n\in{\bigwedge}{^{n-1}}\mathbb{R}^n=\mathbb{R}^n\,,$$
where $\wedge$ is the exterior product.  It can be easily seen that the $(n-1)$-volume of the $(n-1)$-simplex
$$S:=X_1X_2\ldots X_n$$
is equal to
$$v:=\frac{1}{(n-1)!}\,\Big\|(a_2-a_1)\wedge (a_3-a_1)\wedge \ldots \wedge (a_n-a_1)\Big\|\,.$$
With some algebraic manipulations, we get
$$v=\frac{1}{(n-1)!}\,\big\|A_1+A_2+\ldots+A_n\big\|\,.$$
As $A_1,A_2,\ldots,A_n$ are mutually orthogonal elements of ${\bigwedge}{^{n-1}}\mathbb{R}^n=\mathbb{R}^n$, we conclude that
$$v^2=v_1^2+v_2^2+\ldots+v_n^2\,.$$
This result is known as the $n$-Dimensional Pythagorean Theorem.  See also here.
A: 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
$$9 V^2 = A^2 a^2 = B^2 b^2 = C^2 c^2 = D^2 h^2, \\
A^2 + B^2 + C^2 =
D^2 h^2 \left( \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \right) =
D^2.$$
A: The Pythagorean Theorem can be extended to any number of dimensions. There are many ways to show this but the simplest is to show how, for any hypotenuse value, there is an odd leg of another 'triple' with the same value. For example, given $(3,4,5)$, the there is a triple $(5,12,13)$ and, joining the two, we have $3^2+4^2+12^2=13^2$ because $3$ and $4$ replace in the first triple replaces the $5$ in the second.
To find the next triple to join the first, we assume that the hypotenuse is $\space C\space $ and a we derive a parameter $\space k\space $ using $\dfrac{k = (C - 1)}{2}$ and generate the next triple using
$$A=2 k + 1,\quad B=2 k^2 + 2 k,\quad C=2 k^2 + 2 k + 1$$
