Prove that YX line is bisector of an AYB angle

[Berrnie]

We need to prove that <AYX equals <BYX

we know that <ACB=<AYB=<AYX+<BYX because they are on the same arc
the same goes for<CAY=<CBY same arc of circle

<CAB+<ABC=<YAB+<ABY from <ACB=<AYB but it gives this<CAY=<CBY

is there anyhtong we can gain from <AIX=90 degrees or with inscribed circle?

i also tried to for something with other angles but it didn’t do that muchany ideas how to prove it?

 

[lev888]

The problem doesn't describe how those shapes were constructed?

 

[Berrnie]

In triangle ABC, point I is the center of the inscribed circle. Point X lies on segment AB and the condition <AIX = 90◦ is satisfied. The circle circumscribed in triangle BIX intersects the circle circumscribed in triangle ABC in point Y different from B, which lies on the same side of line AB as point C. Show that the line Y X is the bisector of the angle <AYB