circumcenter of CEF lies on circumcircle of trapezoid ABCD

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circumcenter of CEF lies on circumcircle of trapezoid ABCD

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Source: 2016 SPbU finals grades 6-9 p5 v5 - Saint Petersburg State University School Olympiad

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#1 h Dec 31, 2020, 10:22 AM

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Trapezoid $ABCD$ ( $AB \parallel CD$ ) is inscribed in a circle $\omega$ . On the ray $DC$ beyond the point $C$ , point $E$ is marked such that $BC = BE$ . Line $BE$ intersects for the second time circle $\omega$ at the point $F$ lying outside the segment $BE$ . Prove that the center of the circumcircle of triangle $CEF$ lies on $\omega$ .

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#2 h Dec 31, 2020, 12:18 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.167445716287652, xmax = 37.523411430188915, ymin = -21.197084180943115, ymax = 14.850274924959974; /* image dimensions */ /* draw figures */ draw((-4,-1)--(10,-1), linewidth(0.8) + blue); draw((10,-1)--(4.605555946281669,7.616332050957947), linewidth(0.8) + blue); draw((4.605555946281669,7.616332050957947)--(1.3944440537183298,7.616332050957946), linewidth(0.8) + blue); draw((-4,-1)--(1.3944440537183298,7.616332050957946), linewidth(0.8) + blue); draw(circle((10,-1), 10.165687613785025), linewidth(0.8) + red); draw(circle((3,0.6143173072068845), 7.183733038493822), linewidth(0.8) + red); draw((7.509876142632964,-4.977376313375138)--(15.394444053718328,7.616332050957949), linewidth(0.8) + blue); draw((4.605555946281669,7.616332050957947)--(15.394444053718328,7.616332050957949), linewidth(0.8) + blue); draw((10,7.616332050957947)--(10,-1), linewidth(0.8) + blue); draw((10,2.228634614413774)--(4.605555946281669,7.616332050957947), linewidth(0.8) + blue); draw((10,2.228634614413774)--(7.509876142632964,-4.977376313375138), linewidth(0.8) + blue); draw((4.605555946281669,7.616332050957947)--(7.509876142632964,-4.977376313375138), linewidth(0.8) + blue); draw((10,2.228634614413774)--(15.394444053718328,7.616332050957949), linewidth(0.8) + blue); draw((10,2.228634614413774)--(-4,-1), linewidth(0.8) + blue); draw((-4,-1)--(4.605555946281669,7.616332050957947), linewidth(0.8) + blue); draw((7.509876142632964,-4.977376313375138)--(-4,-1), linewidth(0.8) + blue); /* dots and labels */ dot((-4,-1),dotstyle); label("$A$", (-3.840556079408989,-0.6093665325926061), NE * labelscalefactor); dot((10,-1),dotstyle); label("$B$", (10.148534117547094,-0.6093665325926061), NE * labelscalefactor); dot((4.605555946281669,7.616332050957947),dotstyle); label("$C$", (4.756512828693401,7.987702375509804), NE * labelscalefactor); dot((1.3944440537183298,7.616332050957946),dotstyle); label("$D$", (1.551465209444703,7.987702375509804), NE * labelscalefactor); dot((15.394444053718328,7.616332050957949),linewidth(4pt) + dotstyle); label("$E$", (15.540555406400784,7.912289490351012), NE * labelscalefactor); dot((7.509876142632964,-4.977376313375138),linewidth(4pt) + dotstyle); label("$F$", (7.659908907306927,-4.681662331167432), NE * labelscalefactor); dot((10,2.228634614413774),linewidth(4pt) + dotstyle); label("$O$", (10.148534117547094,2.5202682014973066), NE * labelscalefactor); dot((10,7.616332050957947),linewidth(4pt) + dotstyle); label("$G$", (10.148534117547094,7.912289490351012), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Nice problem

$\color{black}\rule{25cm}{1pt}$
We have that $ABCD$ is an isosceles trapezoid.
Let $\alpha = \angle DAB = \angle ABC$ .
Also we have that $\angle BCE=\angle BEC=\alpha$ .
Let $G$ be the midpoint of $CE$ and let $O$ be the second intersection of the circle $\omega$ with $BG$ .
Since $BG$ is the perpendicular bisector of $CE$ we have that $OC=OE$ .

Since we have that $E,B,F$ are colinear we must have that:
$180=\angle ABF+\angle ABE=\angle ABF+\angle ABC+\angle CBE$ since we have that $\angle CBE=180-2\alpha$ from the upper relation we have that $\angle ABF = \alpha$ .
This implies that $AC=AF$ and we have that $\angle COF = \angle CBF = 2\alpha$ .
This gives us $\angle COF = 2 \angle CEF$ and since $B$ is on $EF$ and $OC=OE$ , we have that $O$ is the circumcenter of $CEF$

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