Prove points are coplanar 
I have that $AX = BX, AW=BW, AY = BY, AZ=BZ$
No matter how, I can't imagine that this points necessarily need to be coplanar, but the exercise asks me to do it: prove that $W,X,Y,Z$ are coplanar.
Any help?
 A: Let $M$ be the midpoint of the line segment $AB$. Then, since each of $$\triangle{WAB},\triangle{XAB},\triangle{YAB},\triangle{ZAB}$$ is an isosceles triangle, we have 
$$\angle{AMW}=\angle{AMX}=\angle{AMY}=\angle{AMZ}=90^\circ.$$
Hence, $W,X,Y,Z$ are on the plane which is perpendicular to $AB$ and passes through $M$.
A: Look at the plane perpendicular to AB and through its midpoint.
A: A plane of points $p$ is can be defined by the equation 
$$p \circ N = 1$$
for some vector $N$, where $\circ$ is the dot product (except for a plane through the origin which is $p \circ N = 0$).  
Consider the expression "all points that are equidistant from $A$ and $B$".  Does this expression characterize a plane?
$$\sqrt{(X - A)\circ (X - A)} = \sqrt{(X - B)\circ (X - B)}$$
$$(X - A)\circ (X - A) = (X - B)\circ (X - B)$$
$$X \circ X - 2\cdot X \circ A + A \circ A = X \circ X - 2 \cdot X \circ B + B \circ B$$
$$2 \cdot X \circ A - 2 \cdot X \circ B = A \circ A - B \circ B$$
$$X \circ \left(\frac{2~(A - B)}{A \circ A - B \circ B}\right) = 1$$
So yes, it is a plane equation.
A: Can you show that WZ is perpendicular to AB, and that the midpoint of AB is coplanar with WZ? The same argument will apply to XY and that's it. You will have WZ and XY coplanar with the midpoint of AB and hence are coplanar.
