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?

 A: You want to show that $\angle WZX + \angle ZWY = 90^\circ. $ This means, equivalently, that arcs $WBX$ and $ZDY$ are together half of the full circle, or that they are together the same as the sum of arcs $ZAW$ and $YCX$. Do you see why sums of those pairs of arcs are equal?
A: 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)]

A: 
The points $ \ A \ , \ B \ , \ C \ , \ D \ $ are placed arbitrarily on the circumference of the circle and we take the points $ \ W \ , \ X \ , \ Y \ , \ Z \ $ to be the midpoints of the consecutive arcs on this circle.  As measured from the center of the circle $ \ O \ $ [marked in green], we will set arc measures
$$ \ m( \widehat{AB} )  \ = \ \alpha \ \ , \ \ m( \widehat{BC} )  \ = \ \beta \ \ , \ \ m( \widehat{CD} )  \ = \ \gamma \ \ , \ \ m( \widehat{DA} )  \ = \ \delta \ \ , \ \ $$
from which it follows that
$$ \ m( \widehat{WX} )  \ = \ \frac{\alpha + \beta}{2} \ \ , \ \ m( \widehat{XY} )  \ = \ \frac{\beta + \gamma}{2} \ \ , \ \ m( \widehat{YZ} )  \ = \ \frac{\gamma + \delta}{2} \ \ , \ \ m( \widehat{ZX} )  \ = \ \frac{\delta + \alpha}{2} \ \ , $$
using the radial lines in green.
Turning our attention to the diagonals [in orange] of (cyclic) quadrilateral $ \ WXYZ \ \ , \ $ we may choose any of the four triangles into which these divide the quadrilateral.  So, for instance, applying a reasonably familiar circle chord theorem, we have
$$ m(\angle ZOW) \ \  = \ \frac{\delta + \alpha}{2} \ \ \Rightarrow \ \ m(\angle ZXW) \ \  = \ \ \frac{\delta + \alpha}{4} $$
and
$$ m(\angle XOY) \ \  = \ \ \frac{\beta + \gamma}{2} \ \ \Rightarrow \ \ m(\angle XWY) \ \ = \ \ \frac{\beta + \gamma}{4}  \ \ . $$
In $ \ \triangle WIX \ $ then, we have
$$ m(\angle WIX) \ \  = \ \ 180º \ - \ \frac{\delta + \alpha}{4} \ -  \ \frac{\beta + \gamma}{4} \ \ = \ \  180º \ - \ \frac{\alpha \ + \ \beta \ + \ \gamma \ + \ \delta}{4} \ \ = \ \ 90º \ \ , $$
since $ \ \alpha + \beta + \gamma + \delta \ = \ 360º \ \ . $  The angle $ \ \angle WIZ \ $ and $ \ \angle XIY \ $ are supplementary to $ \ \angle WIX \ \ $ and $ \ \angle ZIY \ $ is its vertical angle, so the diagonals $ \ XZ \ $ and $ \ WY \ $ are orthogonal.
(Note that as quadrilateral $ \ WXYZ \ $ is cyclic, angles at opposite vertices of a diagonal are supplementary, $ \ \triangle WIX \ \sim \ \triangle ZIY \   , \ $ and $ \ \triangle WIZ \ \sim \ \triangle XIY \ \ . \ $ However, in general, no two vertex angles of the quadrilateral need be equal.)
A: For some variety, here is a complex numbers approach.
In the complex plane, let $A, B, C, D$ represent the numbers $e^{2ai}, e^{2bi}, e^{2ci}, e^{2di}$.
So $W, X, Y, Z$ represent the numbers $e^{(a+b)i}, e^{(b+c)i}, e^{(c+d)i}, e^{(a+d+\pi)i}$.
$\text{{gradient of $WY$}}\times\text{{gradient of $XZ$}}$
$=\dfrac{\text{Im}(e^{(c+d)i}-e^{(a+b)i})}{\text{Re}(e^{(c+d)i}-e^{(a+b)i})}\times \dfrac{\text{Im}(e^{(a+d+\pi)i}-e^{(b+c)i})}{\text{Re}(e^{(a+d+\pi)i}-e^{(b+c)i})}$
$=\dfrac{\sin(c+d)-\sin(a+b)}{\cos(c+d)-\cos(a+b)}\times \dfrac{\sin(a+d+\pi)-\sin(b+c)}{\cos(a+d+\pi)-\cos(b+c)}$
$=\dfrac{\sin(c+d)-\sin(a+b)}{\cos(c+d)-\cos(a+b)}\times \dfrac{\sin(a+d)+\sin(b+c)}{\cos(a+d)+\cos(b+c)}$
which equals $-1$ by the sum to product identities.
$\therefore WY\perp XZ$
