Prove that right-angled parallelepiped $ABCDA_1B_1C_1D_1$ is a cube given the following. 
In right-angled parallelepiped $ABCDA_1B_1C_1D_1$ diagonal $AC_1$ is
  perpendicular to the plane containing triangle $A_1BD$. Prove that
  $ABCDA_1B_1C_1D_1$ is a cube.

This is what I've tried:
1)Defined coordinate system
Let the coordinates of  $ABCDA_1B_1C_1D_1$ be $A (0,0,z),\space B(0,y,z),\space C(x,y,z),\space$$D(x,0,z),$ and $A_1(0,0,0),\space B_1(0,y,0),\space C_1(x,y,0),\space D_1(x,0,0)$.
Define $A_1$  to be the origin of our vector system.
Now let $P(x_0,y_0,z_0)$ be the point where $AC_1 \cap A_1BD$,so we have that $\vec P =(x_0 \space \space \space y_0 \space\space\space z_0)$
2)My idea to tackle the problem
Now my idea was to take the cross product of two vectors which lie in $A_1BD$ and then cross this new vector ,say $\vec Z$, with the vector  $\vec{AP}$ which goes in the same direction so that $\vec Z \times \vec{AP} =0$ which would prove $x=y=z$ .
However when I do the computations I am able only to get $x=z$ and I can't figure out how to get $y=z=x$ 
My  computations:
$\vec Z=\vec P \times(\vec D -\vec P)=$$\left(\begin{matrix}
\vec i & \vec j & \vec k \\
x_0-x& y_0 &z_0-z \\
x & 0 & z  \\
\end{matrix}
\right)$$=
\left(
\begin{matrix}
y_0z \\
xz_0-zx_0 \\
-xy_0 \\
\end{matrix}\right)$
Now , $\vec AC_1=\vec A-\vec {C_1}=\left(\begin{matrix}
-x \\
-y \\
z \\
\end{matrix}\right)$
Since $\vec {AC_1}$ and $\vec Z$ go in the same direction I must have $\vec {AC_1} \times \vec Z =0$
So,$\left(\begin{matrix}
-x \\
-y \\
z \\
\end{matrix}\right)$
$\times \left(\begin{matrix}
y_0z \\
xz_0-zx_0 \\
-xy_0 \\
\end{matrix}\right)=0$
From this I have that $$y(xy)-z(xz_0-zx_0)=0 $$
$$ y_0z^2-x^2y_0=0 \tag 2$$
$$-x(xz_0-zx_0)+y(y_0z)=0 $$
From this sistem of equations ,however, I can only get that $x=z$ (this from$(2)$).
Can someone help me ?What do I have to do now?Where is the fall in my thinking (or in my computations) ?
Edit:
@Aretino 
(I am omitting some vector notation just to be faster typing,)
$\vec Z = (B-A_1) \times (D-A_1) =\left(\begin{matrix}
0 \\
y \\
z \\
\end{matrix}\right)\times \left(\begin{matrix} x \\ 0 \\ z \\ \end{matrix}\right)=\left(\begin{matrix} yz \\ xz \\ xy \\ \end{matrix}\right)$
$\vec{Z}\times \vec {AC_1}=\left(\begin{matrix} yz \\ xz \\ xy \\ \end{matrix}\right) \times \left(\begin{matrix} x \\ y \\ -z \\ \end{matrix}\right)=0$ 
From this I have $$-x^2z^2-x^2y^2=0$$
$$y^2z-x^2z=0$$
$$y^2-x^2z=0$$
From which it follows $x=y$ but doesn't follow $x=z$
I think that my mistake lies in the fact that I am stuck with this approach.But why doesn't it lead me to $x=y=z$ conceptually?
Last Edit (I promise):
I found my mistake ,it should have been $\left(\begin{matrix} yz \\ xz \\ **-xy** \\ \end{matrix}\right)$
The algebra definitely killed my soul now.
Thanks for your help both Aretino and Christian Blatter !
 A: $AC_1 \perp (A_1BD)$ so $AC_1 \perp BD$ 
$CC_1 \perp (ABC)$ so $CC_1 \perp BD$ 
now we have that 
$BD \perp (ACC_1)$ so $BD \perp AC$  
and because of this $ABCD$ is a square
now you do something similar for $ABB_1A_1$:
$AC_1 \perp (A_1BD)$ so $AC_1 \perp A_1B$ 
$AD \perp (ABA_1)$ so $AD \perp A_1B$ 
now we have that 
$A_1B \perp (AC_1D)$ so $A_1B \perp AB_1$  
and because of this $ABB_1A_1$ is a square
and now your right-angled parallelepiped is a cube
A: Your idea is fine, just drop point $P$ and define $\vec Z=(\vec B-\vec A_1)\times(\vec D-\vec A_1)$. 
The problem with your solution is that you introduced $P$ without constraining its coordinates so that it lies on plane $A_1BD$.
A: Let $A=(0,0,0)$, and let the parallelepiped have dimensions $x$, $y$, $z$ in the coordinate directions. Then
$$\vec{BD}\times\vec{BA_1}=(-x,y,0)\times(-x,0,z)=(yz,zx,xy)$$
is perpendicular to the plane of triangle $A_1BD$. It follows that necessarily
$$(x,y,z)=\vec{AC_1}=\lambda(yz,zx,xy)$$ for some $\lambda\ne0$. This implies $x^2=\lambda\>xyz$, and similarly for the others; hence $x^2=y^2=z^2$.
