A cyclic quadrilateral $ABCD$ is given. The lines $AB$ and $DC$ meet at point $E$, and the lines $BC$ and $AD$ meet at point $F$. Let $I$ be the incenter of triangle $AED$, and a ray with origin $F$ be perpendicular to the bisector of angle AID. In which ratio this ray dissects the angle $AFB$?
Problem
Source: Sharygin 2018
Tags: geometry
TheDarkPrince
05.04.2018 09:32
Let's assume $\angle A<\angle B$ and $F$ closer to $D$ than $A$. The other cases can be similarly worked out. Let $G$ be a point on the angle bisector of $\angle AID$ such that $FG$ is the perpendicular to the angle bisector. Let $IG$ meet $AD$ at $H$. We claim that $\angle GFA:\angle GFB=1:3$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -7.762304280306645, xmax = 29.577309676190307, ymin = -5.834042735941312, ymax = 14.794563662975825; /* image dimensions */ pair D = (0.,3.3704012810777866), A = (-2.,0.), B = (8.,0.), C = (7.579494983919816,1.433417355015249), F = (4.691793204230134,11.277014194122435), I = (0.7476406172134774,1.5645283172324513), G = (1.2968997016574533,1.3761928219324864), H = (-0.764026178113137,2.0828638763330454); /* draw figures */ draw((13.188513988164724,0.)--D, linewidth(1.)); draw((13.188513988164724,0.)--A, linewidth(1.)); draw(F--A, linewidth(1.)); draw(F--B, linewidth(1.)); draw(D--I, linewidth(1.)); draw(I--A, linewidth(1.)); draw(H--G, linewidth(1.)); draw(G--F, linewidth(1.)); /* dots and labels */ dot(D,dotstyle); label("$D$", (-0.4780189346949447,3.684498198870602), NE * labelscalefactor); dot(A,dotstyle); label("$A$", (-2.6204558010513272,0.07296176701270547), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (7.969303566938792,-0.7840129795298463), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (7.693847398407256,1.7256987782019124), NE * labelscalefactor); dot((13.188513988164724,0.),linewidth(4.pt) + dotstyle); label("$E$", (13.325395732829747,0.2565992127003951), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (4.8168607493001145,11.51969588154536), NE * labelscalefactor); dot(I,linewidth(4.pt) + dotstyle); label("$I$", (0.8686556670147814,1.8175175010457572), NE * labelscalefactor); dot(G,linewidth(4.pt) + dotstyle); label("$G$", G,2*dir(0)); dot(H,linewidth(4.pt) + dotstyle); label("$H$", H,2*dir(180)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We have \[\angle AFB = 180^{\circ}-(\angle DAB+\angle CBA)=180^{\circ}-(\angle A+\angle B).\]We also have $\angle AID = 90^{\circ}+\frac{\angle AED}{2}=90^{\circ}+\frac{\angle B-\angle A}{2}$. So $\angle AIH = \frac{\angle AID}{2}=45^{\circ}+\frac{\angle B-\angle A}{4}$. So, \[\angle HFG = 90^{\circ}-\left(\frac{\angle A}{2}+45^{\circ}+\frac{\angle B-\angle A}{4}\right)=45^{\circ}-\frac{\angle A+\angle B}{4}=\frac{180^{\circ}-(\angle A+\angle B)}{4}.\]$\angle GFB=\angle AFB-\angle GFA$. So, $\angle AFG:\angle BFG=1:3$.