circle

[ertyu]

From point $A $
outside the circle $(O, r) $
fetch two contiguous segments $AB, AC,$ and let $D, E$ The mid-points $AB, AC,$
Let $Z$ the intersection of segment $CD,$ circle $(O, r).$
If $ {\angle A} = {22 }$
find the angle $\ANGLE{ DZE}$

[sopor]

Is this picture right?
http://plasmon.rghost.ru/47258459/image.png
If yes, then the answer is $\arcsin{(\frac{\sin{22^{\circ}}}{\sqrt{32\sin^{4}{11^{\circ}}-8\sin^{2}{11^{\circ}}+1}})} \approx 25.61^{\circ}$. This is confirmed by Wolfram and the picture I uploaded.
My solution is not "olympiad" at all, it is actually an algebraic bash. If someone had found elegant solution, that would have been interesting

[ertyu]

I am very sorry
I made ​​a big mistake $\angle A=36^{\circ}$

[sopor]

In this case the answer is much better: $\arcsin{(\frac{\sin{36^{\circ}}}{\sqrt{32\sin^{4}{18^{\circ}}-8\sin^{2}{18^{\circ}}+1}})} = 54^{\circ}. It is because $\sin{36^{\circ}}$ and $\sin{18^{\circ}}$ have normal radical form, so you can simplify the expression above.