In an orthogonal parallellepipide (cuboid) with sides $a,b,c$, a plane bisects it in halv through three side-diagonals of the cuboid, leaving a tetrahedral with three of it's sides being right triangles. Denote these right triangle sides by $A,B,C$ and denote the fourth side by $D$. Show that $A^2+B^2+C^2=D^2.$
My method starts with putting the cuboid in a coordinate system like this below.
Fixing the vectors $\vec{u}=(a,0,0)-(0,0,-c) = (a,0,c)$ and $\vec{v}=(a,b,-c)-(0,0,-c)=(a,b,0).$
Now, the area of the fourth triangle is $$D=\frac{1}{2}|\vec{u}\times\vec{v}|= \frac{1}{2}\left(\begin{vmatrix} 0 & c \\ b & 0 \end{vmatrix}-\begin{vmatrix} a & c \\ a & 0 \end{vmatrix}+\begin{vmatrix} a & 0 \\ a & b \end{vmatrix}\right)=-\frac{bc}{2}+\frac{ac}{2}+\frac{ab}{2}.$$
Now, by the figure it follows that $A^2=\frac{b^2c^2}{4}, \ B^2=\frac{a^2b^2}{4}$ and $C^2=\frac{a^2c^2}{4}$ so $$A^2+B^2+C^2=\frac{b^2c^2}{4}+\frac{a^2b^2}{4}+\frac{a^2c^2}{4}.$$
But Squaring $D$ does not give me $\frac{b^2c^2}{4}+\frac{a^2b^2}{4}+\frac{a^2c^2}{4}.$ This method should work!
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Note: I don't want to know other methods, I want to know why the above doesn't work.
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The result you want to prove has a name : it is called De Gua's theorem.
In fact :
$$\vec{u}\times\vec{v}=\begin{pmatrix} -bc\\ \ \ ac\\ \ \ ab \end{pmatrix}$$
giving:
$$D=\frac{1}{2}\|\vec{u}\times\vec{v}\|=\frac12\sqrt{(bc)^2+(ac)^2+(ab)^2}$$
instead of
$$-\frac{bc}{2}+\frac{ac}{2}+\frac{ab}{2}.$$