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?



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.)
