let IY cut AC to C' and IX cut AB to B' and Mis midpoint of arc BC. CY and XB is tangent to circumcircle of ABC . so we have <ABX=<C and <YCA=<B . we have AB=BX .so <AXB=90-c/2=A/2 + B/2 , <BIM=B/2+A/2 so AIBX is cyclic. with similar way we can prove AICY is cyclic. then <BXI=<BAI=<A/2 .so <AB'I=<XB'B=<A/2+<C and <AC'I=<YC'C=<A/2+<B. so <AB'I+<AC'I=<A+<B+<C=180 so AB'IC' is cyclic . so <B'IC'=180-<B'AC'. so <XIY=180-<BAC . so <XIY+<BAC=180

tenplusten wrote:

yeah its pretty easy i actually proved this by saying that the midpoint of arc $AB$ is the ccircumcenter of the quadrilateral $AIBX$ instead just the easy angle chasing (i proved it by saying that finding two congruent triangles its pretty easy to find)

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.002609158652927cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ real xmin = -2.3758851496727185, xmax = 2.125419429653745, ymin = -1.4502465613817708, ymax = 1.557802272330799; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen ffqqff = rgb(1.,0.,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pair A = (-0.34138763512824133,0.9399225939307698), B = (-0.8434295555868877,-0.5372397833021165), C = (0.8206832567796607,-0.5713834019653787), X = (-1.6816019232267987,0.7786332579655526), Y = (1.9099826026793774,0.9931873078696632), I = (-0.17458011951149183,-0.06844275845402309), D = (-0.020513293280516515,-0.9997895802611605); filldraw(A--B--C--cycle, invisible, linewidth(0.8) + ffdxqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(0.4) + linetype("0 3 4 3") + xdxdff); draw(A--B, linewidth(0.8) + ffqqff); draw((-0.584761677134383,0.19874245372340676)--(-0.6000555135807463,0.20394035690524648), linewidth(0.8) + ffqqff); draw(B--C, linewidth(0.8) + ffdxqq); draw(C--A, linewidth(0.8) + blue); draw((0.23067167460104915,0.1694436499142817)--(0.25628206249064794,0.18913594676070974), linewidth(0.8) + blue); draw((0.22301355916077187,0.17940324520468126)--(0.2486239470503707,0.1990955420511093), linewidth(0.8) + blue); draw(C--Y, linewidth(0.8) + blue); draw((1.348487154111244,0.21497620652377106)--(1.3750001586704705,0.19651708649637006), linewidth(0.8) + blue); draw((1.3556657007885666,0.22528681940791462)--(1.3821787053477934,0.20682769938051362), linewidth(0.8) + blue); draw(B--X, linewidth(0.8) + ffqqff); draw((-1.2693277011070272,0.1163577183781761)--(-1.2557037777066593,0.1250357562852603), linewidth(0.8) + ffqqff); draw(X--I, linewidth(0.8)); draw(I--Y, linewidth(0.8)); draw(X--Y, linewidth(0.8)); draw(A--D, linewidth(0.8)); draw(D--B, linewidth(0.8)); draw(D--C, linewidth(0.8)); draw(circle((0.7927436701364873,0.6095551439029391), 1.1812690081515589), linewidth(0.4) + linetype("0 3 4 3") + ffxfqq); draw(circle((-0.9468104963674625,0.3217916779041985), 0.8652298366670751), linewidth(0.4) + linetype("0 3 4 3") + ffxfqq); draw(A--X, linewidth(0.8)); draw(A--Y, linewidth(0.8)); draw(I--B, linewidth(0.8)); /* dots and labels */ dot(A,blue); label("$A$", (-0.3800914127464715,1.0193687102104105), NE * labelscalefactor,blue); dot(B,blue); label("$B$", (-0.9185249748668619,-0.6677231177668064), NE * labelscalefactor,blue); dot(C,blue); label("$C$", (0.8726640084536368,-0.6569544465243986), NE * labelscalefactor,blue); dot(X,blue); label("$X$", (-1.7620708888554735,0.839890856170281), NE * labelscalefactor,blue); dot(Y,blue); label("$Y$", (1.9746580322600358,1.0229582672912132), NE * labelscalefactor,blue); dot(I,blue); label("$I$", (-0.1898448874639336,-0.1974911401816672), NE * labelscalefactor,blue); dot(D,blue); label("$D$", (-0.006777476343000859,-1.1128281957863275), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] solution: by a simple angle chase we get that $AYCI$ and $AIBX$ are cyclic because $\angle AIC= 90^{\circ}+\frac{B}{2}$ and $\angle AYC=90^{\circ}-\frac{B}{2}$.the rest is trivial because then $YI||CD$ and $XI||BD$ so $\angle BAC+\angle XIY=\angle BAC+\angle BDC=180^{\circ}$.$\boxed{Q.E.D}$

Claim : $AXBI$ and $AYCI$ are cyclic. Proof : $\angle AXB = \angle 90 - \frac{ABX}{2} = \angle 90 - \frac{C}{2} = \angle 180 - \angle AIB$. we prove the other one with same approach. Now we have $\angle XIY + \angle BAC = \angle XBA + \angle YCA + \angle BAC = \angle ACB + \angle ABC + \angle BAC = \angle 180$. we're Done.