Proof that two non-parallel planes must intersect? I managed to find, by enumeration, the intersection point of two planes $ax+by+cz+d=0$ and $ex+fy+gz+h=0$, in all possible cases (with the condition that the planes are not parallel). But this is a very ugly proof.
I wonder if there is a quicker and more elegant proof (without linear algebra --- this is high school level (Euclidean) geometry)?

It seems that this is not meant to be a problem about the axioms of euclidean geometry, but about 3d analytic geometry. If the two planes are not parallel then the two vectors ${\bf u}:=(a,b,c)$ and ${\bf v}:=(e,f,g)$ have a non-vanishing cross product ${\bf u}\times{\bf v}$. It follows that, e.g., $af-be\ne0$. Therefore the system of linear equations
$$\eqalign{ax+by&=-cz-d\cr ex+fy&=-gz-h\cr}$$
has a unique solution $(x,y)$ for any choice of $z$. In other words, by solving this system we obtain $x$ and $y$ as functions of $z$:
 $$x=Az+B,\quad y=Bz+D\ ,$$ whereby $A$, $\ldots$, $D$ depend on the data $a$, $\ldots$, $h$. We then can use these functions in order to set up a parametric representation of the line $\ell$ (not "the point") of intersection of these two planes as follows:
$$\ell: \quad z\mapsto\bigl(Az+B,Cz+D,\ z\bigr)\qquad(-\infty<z<\infty)\ .$$
Here we have used the $z$-coordinate of the moving point on $\ell$ as parameter.
