Circle $\Gamma$ is the incircle of $\triangle ABC$ and is also the circumcircle of $\triangle XYZ$. The point $X$ is on $\overline{BC}$, poi

Circle \;\;\Gamma\;\; \text{ is the incircle of }\triangle ABC \text{ and is also the circumcircle of }\triangle XYZ\text{ The point X is on }\overline{BC} \text{ point Y is on }\overline{AB}, \textand the point Z is on }\overline{AC}.\;\; If\;\; $\angle A=40^\circ, \;\;\angle B=60^\circ,\;\; and \;\;\angle C=80^\circ, \text{what is the measure of }\angle AYX\; ?

\(\text{Circle }\Gamma\;\; \text{ is the incircle of } \triangle ABC \text{ and is also the circumcircle of } \triangle XYZ\\ \text{ The point X is on }\overline{BC} \text{and the point Y is on }\overline{AB}, \\ \text{and the point Z is on }\overline{AC}.\\ If\;\; \angle A=40^\circ, \;\;\angle B=60^\circ,\;\; and \;\;\angle C=80^\circ, \text{what is the measure of }\angle AYX\; ? \)

   Let O be the centre of the circle.

  OX=OY                                 equal radio

  So

  OXY is an isosceles triangle

  120+2

  so

  < AYX =

I am sorry, this was a full answer but 3/4 of it has been deleted.

There is obviously a software problem.       

I shall report it as a problem :/

See the following image :

Construct angle bisectors of each vertex angle of triangle ABC

Angle AYC  =  180 - angle ACY  - angle YAC  =   180 - 40  - 40   =  100°

Angle AOC   = 180  - angle OAC  - angle OCA  = 180 - 20 - 40  =  120°

And angle  YOX  is a vertical angle to angle AOC  ....so it measures  120°

And since they are equal radii, OY  = OX

So angle OYX  = angle OXY

So triangle OYX is isosceles

And.....angle OYX  = [  180 - 120 ] / 2   = 60  / 2   = 30°

And AYX  =  angle AYC + angle OYX =   100 + 30   = 130°