Please correct the question..... (what is circumcircle lies on I?)

It is the circumcentre of $BDX$ ri8?

yea as bbai said

It is really easy problem. Let $ E $ be the point which is intersection of $w$ and perpendicular bisector of $BD$. We can clearly get by $AB=AX$ and $AE$ is angle bisector of $\angle BAD$ that $AE$ is perpendicular bisector of $BX$. Obviously we have circumcenter of triangle is the intersection of perpendicular bisectors of the sides. This completes solution.

can you add a diagram with a more detailed proof?

I wrote full solution. No I cannot

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(18cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.5484999929, xmax = 32.0617905897, ymin = -8.5605248937, ymax = 12.4482585696; /* image dimensions */ /* draw figures */ draw((6.9470836148,4.6391593419)--(9.5,-1.36)); draw((9.5,-1.36)--(0.36,-1.48)); draw((0.36,-1.48)--(6.9470836148,4.6391593419)); draw(circle((4.90835253601,0.228815174019), 4.85876116832)); draw(circle((6.9470836148,4.6391593419), 6.5197618729)); draw((xmin, 2.53247973902*xmin-12.9541891578)--(xmax, 2.53247973902*xmax-12.9541891578)); /* line */ draw(circle((7.06216865575,-4.12648461051), 3.68733762534)); draw((3.38387569907,-4.38459251059)--(9.5,-1.36)); draw((3.30657716457,1.50297920027)--(9.5,-1.36)); draw((3.38387569907,-4.38459251059)--(0.984195647484,2.00263474148)); draw((xmin, -2.66169952442*xmin + 4.62226782832)--(xmax, -2.66169952442*xmax + 4.62226782832)); /* line */ draw((xmin, -0.462261220708*xmin + 3.03148159673)--(xmax, -0.462261220708*xmax + 3.03148159673)); /* line */ /* dots and labels */ dot((6.9470836148,4.6391593419),dotstyle); label("$A$", (7.10911997813,4.86386020738), NE * labelscalefactor); dot((9.5,-1.36),dotstyle); label("$B$", (9.64989342946,-1.12781449875), NE * labelscalefactor); dot((0.36,-1.48),dotstyle); label("$C$", (0.510693403019,-1.24158047419), NE * labelscalefactor); dot((4.55254719853,-1.42495561665),dotstyle); label("$X$", (4.72003449404,-1.20365848237), NE * labelscalefactor); dot((3.38387569907,-4.38459251059),dotstyle); label("$D$", (3.5444527479,-4.16157384363), NE * labelscalefactor); dot((7.06216865575,-4.12648461051),dotstyle); label("$O$", (7.22288595356,-3.89611990095), NE * labelscalefactor); dot((0.723269313311,2.69714224106),dotstyle); label("$F$", (0.889913321128,2.92983862502), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the circle with center $A$ that passes through $B$ and $X$ intersect $\omega$ at a point $F\neq B$. By a well known lemma (Fact 5), $X$ is the incenter of $\triangle BFD$ then by the same well known lemma (Fact 5) we are done.

A rather simple solution: Let the perpendicular bisector of $BX$ cut $BX$ at $M$ and $\omega$ at $O$. We claim that $O$ is the reqd. circumcentre. Note that since $O$ is on the perpendicular bisector of $BX$, $OB=OX$. Also, $OM$ produced bisects $\angle BAX$. So, $OB=OD$. Thus, $OB=OD=OX$, proving our claim.

$O$ is the circumcenter of $\bigtriangleup BXD$.$\angle BOD=360^{\circ}-2(180^{\circ}-\angle AXB)=2\angle AXB$.Since $\angle BAX=180^{\circ}-2\angle AXB \implies O$ lies on $\omega$.