Proving two chords are perpendicular I am stuck on the following problem: Let A, B, C, and D be placed consecutively on a circle. Let W, X, Y, and Z be the midpoints of AB, BC, CD, and DA, respectively. Show that chords WY and XZ are perpendicular.
I have been trying to relate things to 180 degrees, but I am getting stuck.
Attached to this is my work so far.
Anyone have any tips or tricks?


EDIT 1:
We use a vector approach and begin with the origin at A or O. A circle of unit diameter touching x-axis at A has equation in polar coordinates $ r= \sin \theta$ and in cartesian coordinates for point B with polar angle $\alpha$ we have for
$$ B: (x,y)= (\sin \alpha \cos \alpha, \sin^2 \alpha) $$
Let $\angle{BAC}= \beta, \angle {CAD}=\gamma$ and we mark out arc centers $(W,X,Y,Z)$ coordinates by chasing angles as shown around the circle. Angle subtended by half of an arc is half of that for full arc.
$$ A:(0,0);~B: (x,y)=( \sin \alpha \cos \alpha, \sin^2 \alpha)$$
$$ W: ( \sin \alpha/2 \cos \alpha/2, \sin^2 \alpha/2)$$
$$ C:( \sin (\alpha+\beta) \cos(\alpha+\beta),\sin^2(\alpha+\beta) ) $$
$$ X:( \sin (\alpha+\beta/2) \cos(\alpha+\beta/2),\sin^2(\alpha+\beta/2) ) $$
$$ D:( \sin (\alpha+\beta+\gamma) \cos(\alpha+\beta+\gamma),\sin^2(\alpha+\beta+\gamma) 
) $$
$$ Y:( \sin (\alpha+\beta+\gamma/2) \cos(\alpha+\beta+\gamma/2),\sin^2(\alpha+\beta+\gamma/2) $$
$$ Z:( \sin (\alpha/2+\beta/2+\gamma/2+\pi/2) \cos(\alpha/2+\beta/2+\gamma/2+\pi/2),\sin ^2(\alpha/2+\beta/2+\gamma/2+\pi/2)) $$

We need to show that vector $\vec {ZX}$ is perpendicular to $\vec{YW}.$
This has been checked in effect analytically by obtaining exact zero dot product of vectors connecting alternate arcs midpoints among the four arcs marked red... for arbitrary $ (\alpha,\beta,\gamma)$. To reduce tedium of trig simplifications, coded the above on Mathematica.
$$ (\vec Z-\vec X)\cdot(\vec Y-\vec W)=0~; $$
Aliter
In hindsight the center of the circle can also be taken as origin, that is more simple. For arbitrary $(al,bt,gm): $
> Clear["`*"]; A[0,0]; B={Cos[al],Sin[al]};W={Cos[al/2],Sin[al/2]}; X={
> Cos[al+bt/2],Sin[al+bt/2]};CC={ Cos[al+bt],Sin[al+bt]};
> Y={Cos[al+bt+gm/2],Sin[al+bt+gm/2]}; 
> Z={Cos[al/2+bt/2+gm/2+Pi],Sin[al/2+bt/2+gm/2+Pi]}; X-Z ;Y-W;
> Simplify[(X-Z).(Y-W)]

