Quadrilateral $ABCD (\overline{AD} < \overline{BC})$ is inscribed in a circle, and $H(\neq A, B)$ is a point on segment $AB.$ The circumcircle of triangle $BCH$ meets $BD$ at $E(\neq B)$ and line $HE$ meets $AD$ at $F$. The circle passes through $C$ and tangent to line $BD$ at $E$ meets $EF$ at $G(\neq E).$ Prove that $\angle DFG = \angle FCG.$
Problem
Source: KJMO 2023 P2
Tags: geometry, circumcircle
bin_sherlo
04.11.2023 22:51
Isn't $AB=(ABH)$? Maybe there's a typo.
soryn
04.11.2023 23:15
The statement is wrong..
jason02
11.11.2023 08:03
ABH should be BCH
pokmui9909
11.11.2023 08:50
jason02 wrote: ABH should be BCH I fixed it. Thanks.
bin_sherlo
11.11.2023 10:41
$(BCH)$ meets $AB$ at $B,H,E$. I think there is a typo again.
pokmui9909
11.11.2023 11:15
bin_sherlo wrote: $(BCH)$ meets $AB$ at $B,H,E$. I think there is a typo again. Fixed. For clarification, I will post a diagram. [asy][asy] import graph; size(9cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -6, xmax = 4, ymin = -6, ymax = 6; pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((1.9839104979522297,5.131636755897212),0.6,-173.78658573705306,-141.07335617819544)--(1.9839104979522297,5.131636755897212)--cycle, linewidth(0.7) + qqwuqq); draw(arc((2.8550116893164565,-0.29599222233968614),0.6,99.11786526732458,131.83109482618215)--(2.8550116893164565,-0.29599222233968614)--cycle, linewidth(0.7) + qqwuqq); draw(circle((-1.28,0.84), 4.288216412449353), linewidth(0.7)); draw((-3.447060457965863,4.540357951809633)--(-4.351347474964859,-2.152594975606787), linewidth(0.7)); draw((-4.351347474964859,-2.152594975606787)--(2.8550116893164565,-0.29599222233968614), linewidth(0.7)); draw((2.8550116893164565,-0.29599222233968614)--(0.04,4.92), linewidth(0.7)); draw((0.04,4.92)--(-3.447060457965863,4.540357951809633), linewidth(0.7)); draw(circle((-0.7010698736103825,-1.4071034336477477), 3.7256253442989484), linewidth(0.7)); draw((0.04,4.92)--(-4.351347474964859,-2.152594975606787), linewidth(0.7)); draw(circle((0.5383898642512908,0.8352558031195925), 2.5780726862277907), linewidth(0.7)); draw((1.9839104979522297,5.131636755897212)--(2.8550116893164565,-0.29599222233968614), linewidth(0.7)); draw((2.8550116893164565,-0.29599222233968614)--(-0.3302119536081377,3.262597900452614), linewidth(0.7)); draw((0.04,4.92)--(1.9839104979522297,5.131636755897212), linewidth(0.7)); draw((1.9839104979522297,5.131636755897212)--(-1.6518415011593357,2.195162175428098), linewidth(0.7)); draw((-4.0226307634221445,0.28035544328986634)--(-1.6518415011593357,2.195162175428098), linewidth(0.7)); dot((0.04,4.92),dotstyle); label("$D$", (0.12,5.12), NE * labelscalefactor); dot((-3.447060457965863,4.540357951809633),dotstyle); label("$A$", (-3.9,4.7), NE * labelscalefactor); dot((-4.351347474964859,-2.152594975606787),dotstyle); label("$B$", (-4.92,-2.56), NE * labelscalefactor); dot((2.8550116893164565,-0.29599222233968614),dotstyle); label("$C$", (3.18,-0.52), NE * labelscalefactor); dot((-4.0226307634221445,0.28035544328986634),dotstyle); label("$H$", (-4.5,0.38), NE * labelscalefactor); dot((-1.6518415011593357,2.195162175428098),linewidth(4pt) + dotstyle); label("$E$", (-1.95,2.3), NE * labelscalefactor); dot((1.9839104979522297,5.131636755897212),linewidth(4pt) + dotstyle); label("$F$", (2.06,5.3), NE * labelscalefactor); dot((-0.3302119536081377,3.262597900452614),linewidth(4pt) + dotstyle); label("$G$", (-0.62,3.48), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]
bin_sherlo
11.11.2023 15:31
$\angle FEC=\angle HBC=\angle ABC=\angle FDC \implies E,D,F,C$ are cyclic. $\angle ECG=\angle GED=\angle FED=\angle FCD \implies \angle GCF=\angle ECD=\angle EFD$
Lufin
06.10.2024 16:57
could use miquel's theorem in square AHED