Here's the full problems and solution link: http://igo-official.ir/wp-content/uploads/2016/12/igo_2016_-_problems__solutions_-_english.pdf

Let us translate the problem like this; Let $ABC$ be a triangle and let $W_1$ and $W_2$ be circles that pass through $A$ and $B$ and $A$ and $C$ respectively,also tanget to $AC,AB$ at $A$ respectively.Let $W_1nW_2=Q$, and let the angle bisector of $A$ intersect $W_1,W_2$ at $E,F$ -respectively.Let the line perpendicular to the internal angle bisector of $A$ ,that pass through $A$ intersect $W_1,W_2$ at $X,Y$.Prove that the perpendicular bisector of $XY$ is tangent to $BEF$. Claim1:.$Q$ is the dumpty point. Its clear from the definition of $W_1,W_2$ that $\angle ABQ=\angle QAC$ and $\angle ACQ=\angle QAC$,and we get that $Q$ is dumpty point,also we know that $OQ \perp AQ$. Claim.2: $O-E-O_1-X$-collinear. Since $E$ lies on the angle bisector,and from the definition of $W_1$ we get that $EA=EB$, also from the definition of $X$ we get that $E-O_1-X$ -colinear,but we know that $OO_1\perp AB$ , wich gives us that $O-E-O_1-X$-collinear. Similarly $F-O-O_2-Y$-collinear. Claim.3: $OX=OY$ this is a very easy angle chase using properties of cyclic quadrilaterals and using the fact of the angle bisector. Claim.4:.$(QEOF)$-cyclic. $\angle ABQ=\angle QEF=\angle QAC$ , also since we have $OQ \perp AQ$., we get that $\angle ABQ=\angle QEF=\angle QAC=\angle QOF$. From an very easy angle chase we get that $OE=OF$ and we know that the perpendicular bisector of $XY$ passes through $O$ and parralel to $EF$ ,we get the desired claim. $\blacksquare$

Here is my solution which uses only angle chasing and some good observations. Solution. WLOG let $F$ be closer to $A$ than $E$. Let $K=EX\cap YF$. Claim:$KFBE$ is cyclic, and $K$ is the midpoint of $\overarc{EF}$ without including $B$. Proof. From the tangency condition we have that: $$\angle KXY=\angle EAD=\angle EAC=\angle KYX\implies KX=KY$$$$\angle EBF+\angle EKF=(\angle ABE-\angle ABF)+2\angle KXY=(180^{\circ}-\angle KXY-\angle KXY)+2\angle KXY=180^{\circ}\implies K\in (FBE)$$$\angle KFE=90^{\circ}-\angle KYX=90^{\circ}-\angle KXY=\angle KEF\implies KE=KF\implies$ $K$ is the midpoint of $\overarc{EF}$.$\square$ Now, the perpendicular bisector of $XY$ goes through $K$ and it's parallel to $EF$.Thus, it is tangent to $(BEF)$ at $K$.$\blacksquare$ [asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ real xmin = -27.49, xmax = 10.69, ymin = -5.04, ymax = 12.2; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen qqzzcc = rgb(0.,0.6,0.8); pen ffqqff = rgb(1.,0.,1.); draw(arc((-14.792492459939695,7.824354341936347),0.6,-61.511757783330864,-10.196172876866019)--(-14.792492459939695,7.824354341936347)--cycle, linewidth(0.8) + ffqqff); draw(arc((-4.164199308615807,5.912755469062378),0.6,169.803827123134,221.1194120295988)--(-4.164199308615807,5.912755469062378)--cycle, linewidth(0.8) + ffqqff); draw(arc((-8.43,6.68),0.6,-100.19617287686604,-48.88058835225206)--(-8.43,6.68)--cycle, linewidth(0.8) + ffqqff); draw(arc((-8.43,6.68),0.5,-151.51175740148,-100.19617287686604)--(-8.43,6.68)--cycle, linewidth(0.8) + ffqqff); draw(arc((-8.03,0.4),0.6,93.64448951277159,144.96007403738557)--(-8.03,0.4)--cycle, linewidth(0.8) + ffqqff); /* draw figures */ draw(circle((-12.325838330571038,3.2791185776706335), 5.1714167059609055), linewidth(0.8) + qqwuqq); draw(circle((-6.776205522958496,3.632598374333854), 3.467231293179855), linewidth(0.8) + qqwuqq); draw((-14.792492459939695,7.824354341936347)--(-4.164199308615807,5.912755469062378), linewidth(0.8)); draw((-8.43,6.68)--(-3.980054775966594,1.5824193966584217), linewidth(0.8) + xfqqff); draw((-8.43,6.68)--(-17.300463455373507,1.8660879288963326), linewidth(0.8) + xfqqff); draw((-8.43,6.68)--(-9.859184201202385,-1.2661171865950762), linewidth(0.8)); draw(circle((-9.330291457922188,-0.009609884826263291), 1.3632821180125532), linewidth(0.8) + red); draw((-9.478345884277749,6.868554905499361)--(-10.672043869835095,0.2317168613472566), linewidth(0.8) + qqzzcc); draw((-10.672043869835095,0.2317168613472566)--(-9.859184201202385,-1.2661171865950762), linewidth(0.8)); draw((-9.859184201202385,-1.2661171865950762)--(-8.03,0.4), linewidth(0.8)); draw((-8.03,0.4)--(-8.43,6.68), linewidth(0.8)); draw((-9.388211737301184,1.3524412796053271)--(-8.03,0.4), linewidth(0.8)); draw((-10.672043869835095,0.2317168613472566)--(-14.792492459939695,7.824354341936347), linewidth(0.8)); draw((-4.164199308615807,5.912755469062378)--(-10.672043869835095,0.2317168613472566), linewidth(0.8)); /* dots and labels */ dot((-8.43,6.68),linewidth(4.pt) + ududff); label("$A$", (-8.49,6.98), NE * labelscalefactor,ududff); dot((-8.03,0.4),linewidth(4.pt) + ududff); label("$B$", (-7.87,-0.06), NE * labelscalefactor,ududff); dot((-3.980054775966594,1.5824193966584217),linewidth(4.pt) + uuuuuu); label("$C$", (-3.79,1.26), NE * labelscalefactor,uuuuuu); dot((-17.300463455373507,1.8660879288963326),linewidth(4.pt) + uuuuuu); label("$D$", (-17.83,1.58), NE * labelscalefactor,uuuuuu); dot((-9.859184201202385,-1.2661171865950762),linewidth(4.pt) + uuuuuu); label("$E$", (-9.97,-1.76), NE * labelscalefactor,uuuuuu); dot((-9.388211737301184,1.3524412796053271),linewidth(4.pt) + uuuuuu); label("$F$", (-9.91,1.44), NE * labelscalefactor,uuuuuu); dot((-14.792492459939695,7.824354341936347),linewidth(4.pt) + uuuuuu); label("$X$", (-15.11,8.04), NE * labelscalefactor,uuuuuu); dot((-4.164199308615807,5.912755469062378),linewidth(4.pt) + uuuuuu); label("$Y$", (-4.01,5.96), NE * labelscalefactor,uuuuuu); dot((-9.478345884277749,6.868554905499361),linewidth(4.pt) + uuuuuu); dot((-10.672043869835095,0.2317168613472566),linewidth(4.pt) + uuuuuu); label("$K$", (-11.13,0.1), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Let M be midpoint of XY and S be where XE meets YF. ∠AYF = ∠FAD = ∠FAC = ∠AXE so SXY is isosceles and MS is perpendicular bisector of XY. Let's prove it touches EFB at S. ∠SFE = ∠YFA = 90 - ∠FYA = 90 - ∠EXA = ∠SEF ---> SEF is isosceles we had MS || AF so ∠FSM = ∠FES so MS is tangent to SEF. let's prove B lies on this circle. first we will prove XYBS is cyclic. ∠SYB = ∠FAB = ∠SXB ---> XYBS is cyclic. ∠BSF = ∠BXF = ∠BEA ---> SFBE is cyclic. we're Done.

Note Labelling are different in this solution

Nice problem! We start by constructing the circumcenter of $(AEF) = K$ , Claim 1:$I-A-G-K , J-C-K-H$

Proof

Claim 2:$DGKH$ is cyclic

Proof

Claim 3 : $KG=KH$

Proof

BY claim 3 we are done.