$\triangle ABC$ has circumcenter $O$; $BO$ and $CO$ meet $AC$ and $AB$ at $D$ and $E$. If $\angle A=\angle EDA=\angle BDE$, show they are $50^\circ$ 
There is surely a purely geometrical solution to this problem, but none forthcoming so far!
Trigonometry confirms the result, however the quest is to solve this geometrically, almost certainly involving a clever construction, leading to an equilateral triangle and hence exposing angle values. So a possible nod to “Adventitious Angles”….

… OR, could it be possible to prove the impossibility of a purely geometrical proof?
$O$ is the circumcenter of $\triangle ABC$. $BO$ cuts $AC$ at $D$, and $CO$ cuts $AB$ at $E$, as in the figure. Angles $EAD$, $EDA$, and $BDE$ are equal. Prove that their value is $50^\circ$

You can quickly work out all the angles in terms of x and y=angle OBC and discover that x+y=90. Also that triangle COD is isosceles and further that it’s sufficient to prove triangle BCE isosceles to get the answer.
This may help ... repeat “may”! Define point F on AD such that CE = CF. Then since triangle COD is isosceles, then (among other things) EF is parallel to OD.  Angle BCF will ultimately be shown to be 60 degrees, so we need to show that triangle BFC is equilateral.  Worth exploring a bit; I’m convinced the key is the equilateral triangle because this forces another relationship between x and y.
UPDATE:-
I have made a bit of progress via the construction of the line BF, as described above.
Since triangle COD is isosceles and CF=CE by construction, then triangle CEF is also
isosceles and is similar to triangle COD. The following can then be proved:-

*

*Angle EFD = 180-2x

*Angle FED = x

*FE=FD

*FD=EO

*triangle EFA is congruent to triangle EDO

*triangle EFA is similar to triangle ABD

Now construct the line FO and drop a perpendicular from O to point G on BC. Then the
following can be proved:

*

*FO bisects angle EOD into x + x and also bisects angle FB into (180-3x) + (180-
3x)

*The points F O and G are colinear

*BG = GC

*Triangles CFG and GFB are congruent

*CF = FB

*Triangle CFB is isosceles with CF = FB

BUT .... we still need to prove that triangle CFB is equilateral, which is very elusive.  There are two distinct ideas here…
SO ... any help is greatly appreciated! Thanks for your interest.
 A: Let $AO$ cut $BC$ at $F$. It is well-known that the quadruple $(DA, DB; DE, DF)$ is harmonic. Since $DE$ is the internal angle bisector of the angle $ADB$, it follows that $DF$ is the external angle bisector of $ADB$. In particular $$\angle ODF = \frac\pi2 -\frac12\angle ADB = \frac\pi2-\angle A=\angle OCF.$$ It follows that the quadrilateral $CFOD$ is cyclic. From this we get $$\pi=\angle C +\angle FOD =\angle C + 2\angle C=3\angle C,$$ hence $\angle C=\frac\pi3$.
On the other hand we have $$\pi-3\angle A=\angle DBA=\frac\pi2-\angle C=\frac\pi2-\frac\pi3=\frac\pi6.$$
Hence $$\angle A=\frac13\left(\pi-\frac\pi6\right)=\frac{5\pi}{18}.$$

A: Not an answer but something which might prove useful in one.
This looks like an adventitious-angles problem, but with the four initial constraints given as two angle-equalities ($\angle EDA=\angle ODE=\angle CAB$) and two length-equalities ($OA=OB=OC$) rather than explicit angles.
Draw diameter $AOG$. Then $OG=OB$. It can be found from angle-chasing that $\angle COG=\angle EOA=8x-360^\circ$, $\angle GOB=\angle AOD=360^\circ-6x$ and $\angle OBG=\angle BGO=3x-90^\circ$.
Now if it could be proved that $BG=OB$, too, then $x=50^\circ$ follows immediately.
