Let $ABCD$ be a cyclic quadrilateral. Circle with diameter $AB$ intersects $CA$, $CB$, $DA$, and $DB$ in $E$, $F$, $G$, and $H$, respectively (all different from $A$ and $B$). The lines $EF$ and $GH$ intersect in $I$. Prove that the bisector of $\angle GIF$ and the line $CD$ are perpendicular.
Problem
Source: MDA TST 2016 P11!
Tags: geometry, cyclic quadrilateral
AlastorMoody
26.03.2019 18:42
Lemma: Consider an isosceles trapezium $WXYZ$ where, $XY,WZ$ are parallel & $XW,YZ$ are non-parallel, then, if $XZ \cap WY=T$, then, $\Delta WTZ$, $\Delta XTY$ are isosceles triangles $--- (\text{Well-Known})$ [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(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.001550312837889, xmax = 7.682910659742747, ymin = -3.8150149612952142, ymax = 6.099931108476055; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-5.06,-2.35)--(-5.2,4.55)--(2,2.99)--cycle, linewidth(0.8) + rvwvcq); draw((-5.06,-2.35)--(-5.2,4.55)--(2,2.99)--(1.4350316556431144,-1.6211944530338902)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((2,2.99)--(-5.06,-2.35), linewidth(0.8) + rvwvcq); draw(circle((-2.1654675141928283,1.1601499344946384), 4.549667116836765), linewidth(0.8)); draw((-5.06,-2.35)--(-5.2,4.55), linewidth(0.8) + rvwvcq); draw((-5.2,4.55)--(2,2.99), linewidth(0.8) + rvwvcq); draw((2,2.99)--(1.4350316556431144,-1.6211944530338902), linewidth(0.8) + rvwvcq); draw((1.4350316556431144,-1.6211944530338902)--(-5.06,-2.35), linewidth(0.8) + rvwvcq); draw((-5.2,4.55)--(1.4350316556431144,-1.6211944530338902), linewidth(0.4)); draw(shift((-5.13,1.1))*xscale(3.450710071854777)*yscale(3.450710071854777)*arc((0,0),1,-88.83763642649726,91.16236357350274), linewidth(0.8)); draw((-1.8293052506916856,0.09361330896691222)--(-3.6383125497479463,4.211634385778722), linewidth(0.4)); draw((-4.433639591285704,-2.279716287083171)--(-1.6839758931079503,1.2797716738500586), linewidth(0.4)); draw((xmin, 8.161863401891818*xmin + 17.941518355037953)--(xmax, 8.161863401891818*xmax + 17.941518355037953), linewidth(0.4) + linetype("4 4") + dtsfsf); /* line */ draw((xmin, -0.1225210409387901*xmin + 0.471467059984009)--(xmax, -0.1225210409387901*xmax + 0.471467059984009), linewidth(0.4)); /* line */ draw((-4.433639591285704,-2.279716287083171)--(-3.6383125497479463,4.211634385778722), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((xmin, 8.161863401891782*xmin + 15.024152885475855)--(xmax, 8.161863401891782*xmax + 15.024152885475855), linewidth(0.4) + linetype("4 4") + dtsfsf); /* line */ draw((-4.433639591285704,-2.279716287083171)--(-1.8293052506916856,0.09361330896691222), linewidth(0.4)); draw((-1.6839758931079503,1.2797716738500586)--(-3.6383125497479463,4.211634385778722), linewidth(0.4)); draw((-1.8293052506916856,0.09361330896691222)--(-5.2,4.55), linewidth(0.4)); draw((-5.06,-2.35)--(-1.6839758931079503,1.2797716738500586), linewidth(0.4)); draw((-4.433639591285704,-2.279716287083171)--(-5.2,4.55), linewidth(0.4)); /* dots and labels */ dot((-5.06,-2.35),dotstyle); label("$A$", (-5.514224453814887,-2.92608876193641), NE * labelscalefactor); dot((-5.2,4.55),dotstyle); label("$B$", (-5.78774020746375,5.005868093880604), NE * labelscalefactor); dot((2,2.99),dotstyle); label("$C$", (2.417732402002139,3.2006641197981116), NE * labelscalefactor); dot((1.4350316556431144,-1.6211944530338902),dotstyle); label("$D$", (1.7339430178799817,-1.9961351995302772), NE * labelscalefactor); dot((-1.8293052506916856,0.09361330896691222),dotstyle); label("$E$", (-1.4935428751766011,-0.34136488995465863), NE * labelscalefactor); dot((-3.6383125497479463,4.211634385778722),dotstyle); label("$F$", (-3.421828938401085,4.787055490961515), NE * labelscalefactor); dot((-4.433639591285704,-2.279716287083171),dotstyle); label("$G$", (-4.091942534840799,-2.803006672794422), NE * labelscalefactor); dot((-1.6839758931079503,1.2797716738500586),dotstyle); label("$H$", (-1.2747302722575107,1.5048664471751636), NE * labelscalefactor); dot((-2.108792924279692,0.729838564191112),dotstyle); label("$I$", (-2.9295005818331314,0.4655065833094859), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By Reim's Theorem, we have, $FG||HE||CD$, hence, $FHEG$ is an isosceles trapezium, hence by a corollary of our (Lemma), angle bisector of $\angle GIF$ is perpendicular to $FG$ $\implies$ and since, $FG||CD$ solves the problem