Are these planes always perpendicular? The picture shows a pyramid (not necessarily a right pyramid).

$V$ is the apex of the pyramid , and $ABCD$ is it's base.
Let $\alpha$ be the plane $AVC$ and $\beta$ the plane $BVD$.
True or false: If $ABCD$ is a square, then $\alpha$ is always perpendicular to $\beta$.
Prove it (using only euclidian geometry; vectors/coordinates are not allowed).

My thought process so far:
I first thought since $ABCD$ is a square, then $AC$ is perpendicular to $BD$. But this doesn't guarantee $\alpha$ is perpendicular to $\beta$. There's no theorem saying that if a line in a plane is perpendicular to a line in another plane, then the planes are perpendicular.
If $V$ were guaranteed to be straight over the center of $ABCD$, then it would be easy to prove the proposition is true. The intersection line between $\alpha$ and $\beta$ would be perpendicular to the base and, since $AC$ is perpendicular to $BD$, the planes would be perpendicular to each other by definition of angle between planes.
I think I'm stuck because the pyramid could be oblique.
Let's just use 3D coordinates and see what happens.
WLOG, let $V = (x_0,y_0,z_0)$ s.t. $z_0 &gt; 0$,
and let $A = (-1,0,0), B = (0,-1,0), C = (1,0,0) , D = (0,1,0)$.
The equation of plane $AVC$ is:
$\begin{bmatrix}
    x       &amp; y &amp; z  &amp; 1 \\
    -1       &amp; 0 &amp; 0  &amp; 1 \\
     1       &amp; 0 &amp; 0  &amp; 1 \\                                                   
     x_0      &amp; y_0 &amp; z_0  &amp; 1
\end{bmatrix} = 0$
Hence it's normal is along the vector $(0, -2z_0, 2y_0)$.
The equation of plane $BVD$ is:
$\begin{bmatrix}
    x       &amp; y &amp; z  &amp; 1 \\
    0       &amp; -1 &amp; 0  &amp; 1 \\
     0       &amp; 1 &amp; 0  &amp; 1 \\                                                   
     x_0      &amp; y_0 &amp; z_0  &amp; 1
\end{bmatrix} = 0$
Hence it's normal is along the vector $(2z_0, 0, -2x_0)$.
Hence the dot product of the two normals is $0$ iff $x_0y_0 = 0$.
So no, in general. Also we can see that if the base is a square, then those planes are perpendicular only when the apex's projection lies on one of the diagonals of the square.
The planes are not always perpendicular.
Imagine a point $E$, halfway between $A$ and $B$. The point $V$ can relocate anywhere along the line from $V$ to $E$ and, provided that it is not coincident with $E$, the volume would still be a pyramid with a square base.
Now take the extreme case where $V$ is very, very close to $E$, but not exactly at $E$. To convey the idea graphically, imagine the distance between $V$ and $E$ is the width of an atom... Technically, we still have a square-based pyramid, but for practical purposes it's a flat surface, like a sheet of paper.
In this scenario, the planes you are asking about, $AVC$ and $BVD$, will also lie on this sheet of paper, and would therefore be parallel (or very nearly so).