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.
