Let $\vec{v}, \vec{w} \in \mathbb{R}^3$. The length of their vector product, i.e. $|\vec{v} \times \vec{w}|$ equals the area of the parallelogram induced by $\vec{v}$ and $\vec{w}$. This can be proven using some elementary, yet (for high school level) cumbersome algebra.
I want to prove it differently: If $\vec{v}$ and $\vec{w}$ have e.g. $v_z = w_z = 0$, then the above computation is very easy and for the area, we get $v_x w_y - v_y w_x =: A_{xy}$. This is what happens if we project the parallelogram onto the $xy$-plane.
Since this computation was so easy, I would like to now finish the proof by seeing that the area of the original parallelogram is $A = \sqrt{A_{xy}^2 + A_{xz}^2 + A_{yz}^2}$.
How can I prove the last equality?
I know the area and the projected area are related to each other via a factor of $\cos(\varphi)$, where $\varphi$ is the angle between the normal of the parallelogram and the normal of the projection plane, but applying this doesn't seem to help.
This is essentially de Gua's theorem, applied to a triangle that's half of the parallelogram.
One proof of de Gua's theorem is to take a right tetrahedron with legs $a,b,c$ and apply Heron's formula to the area of the face with sides $\sqrt{a^2+b^2}, \sqrt{a^2+c^2}, \sqrt{b^2+c^2}$, but this doesn't let us get out of doing lots of algebra.
Maybe there are clever proofs out there.