let $ABCD$ be a cyclic quadrilateral with $E$,an interior point such that $AB=AD=AE=BC$. Let $DE$ meet the circumcircle of $BEC$ again at $F$. Suppose a common tangent to the circumcircle of $BEC$ and $DEC$ touch the circles at $F$ and $G$ respectively. Show that $GE$ is the external angle bisector of angle $BEF$
Problem
Source: Nigerian Mathematics olympiad round 2 2021 problem 5
Tags: geometry, cyclic quadrilateral, circumcircle, angle bisector
MiltonMath12
16.02.2021 11:23
[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 = -23.7235899320119, xmax = 17.841374957763243, ymin = -13.414452670362879, ymax = 18.948765132582974; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ffxfqq = rgb(1,0.4980392156862745,0); draw((-8.943952945265764,-3.7535559967871945)--(0.6093359192549261,-3.7697494012158588)--(-0.34956785785916633,5.735306827000697)--(-7.952831476658529,5.748194820287932)--cycle, linewidth(0)); /* draw figures */ draw(circle((-4.16,0.55), 6.434811574552903), linewidth(0.4) + wrwrwr); draw(circle((-4.142478737339204,10.886658002157972), 6.397076822226873), linewidth(0.4) + linetype("4 4") + ffxfqq); draw(circle((-0.8342466330571188,0.885513792737428), 4.8739517837496), linewidth(0.4) + linetype("0 3 4 3") + red); draw((xmin, -0.32796475572505346*xmin + 3.139946387723101)--(xmax, -0.32796475572505346*xmax + 3.139946387723101), linewidth(0.4) + wrwrwr); /* line */ draw((xmin, -1.7481319920042995*xmin-2.704549786888974)--(xmax, -1.7481319920042995*xmax-2.704549786888974), linewidth(0.4) + wrwrwr); /* line */ draw((3.944040352112478,1.8464401570727782)--(-0.34956785785916633,5.735306827000697), linewidth(0.4) + wrwrwr); draw((-0.34956785785916633,5.735306827000697)--(-4.1153577024663015,4.489638671333679), linewidth(0.4) + wrwrwr); draw((-9.642996084523405,14.152680167238586)--(-7.952831476658529,5.748194820287932), linewidth(0.4) + wrwrwr); draw((xmin, 0.33078536165551004*xmin + 5.850938757285784)--(xmax, 0.33078536165551004*xmax + 5.850938757285784), linewidth(0.4) + wrwrwr); /* line */ draw((-0.34956785785916633,5.735306827000697)--(-9.642996084523405,14.152680167238586), linewidth(0.4) + wrwrwr); draw((0.6093359192549261,-3.7697494012158588)--(3.944040352112478,1.8464401570727782), linewidth(0.4) + wrwrwr); draw((xmin, -5.677119115188735*xmin + 24.245384855454077)--(xmax, -5.677119115188735*xmax + 24.245384855454077), linewidth(0.4) + wrwrwr); /* line */ /* dots and labels */ dot((-8.943952945265764,-3.7535559967871945),dotstyle); label("$A$", (-8.776361737308743,-3.3149741604283207), NE * labelscalefactor); dot((0.6093359192549261,-3.7697494012158588),dotstyle); label("$B$", (0.7844779187626436,-3.3149741604283207), NE * labelscalefactor); dot((-0.34956785785916633,5.735306827000697),linewidth(4pt) + dotstyle); label("$C$", (-0.15814007549791578,6.111205782177267), NE * labelscalefactor); dot((-7.952831476658529,5.748194820287932),linewidth(4pt) + dotstyle); label("$D$", (-7.7888571718929205,6.111205782177267), NE * labelscalefactor); dot((-4.1153577024663015,4.489638671333679),dotstyle); label("$E$", (-3.9286120525401533,4.944154932140385), NE * labelscalefactor); dot((3.944040352112478,1.8464401570727782),linewidth(4pt) + dotstyle); label("$F$", (4.106084184252234,2.2060740916692376), NE * labelscalefactor); dot((-9.642996084523405,14.152680167238586),linewidth(4pt) + dotstyle); label("$K$", (-9.449660304637716,14.504994588211767), NE * labelscalefactor); dot((-5.191663806046351,-1.2981348902824492),linewidth(4pt) + dotstyle); dot((3.9658076202594272,1.7310226073181079),linewidth(4pt) + dotstyle); dot((2.1576073471603885,11.996390941818188),linewidth(4pt) + dotstyle); label("$G$", (2.355507909196909,12.35043917275906), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $\angle BEF+90-\frac{\angle DAE}{2}+90-\frac{\angle EAB}{2}=180$. Then $\angle BEF=\frac{\angle DAB}{2}$. Let $K=BE\cap \odot(\triangle CED)$. $\angle KED=\frac{\angle DAB}{2}=\angle DCK$ then $K,C,F$ are collinear. To finish, $\angle CGF=\angle CEG$ and $\angle CEF=\angle CFG$ then $\angle GEF=\angle CGF+\angle CFG$. And $\angle KCG=\angle CGF+\angle CFG$ then $\angle KCG=\angle GEF$ but $\angle KGC=\angle KEG$ then $\angle KEG=\angle GEF$ and we finished