Quadrilateral $ABCD$ is given such that $$\angle DAC = \angle CAB = 60^\circ,$$and $$AB = BD - AC.$$Lines $AB$ and $CD$ intersect each other at point $E$. Prove that \[ \angle ADB = 2\angle BEC. \] Proposed by Iman Maghsoudi
Problem
Source: 6th Iranian Geometry Olympiad (Elementary) P4
Tags: IGO, Iran, geometry
GeoMetrix
21.09.2019 05:37
The one i wrote on the contest [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -21.813738705649882, xmax = 25.45308380311172, ymin = -15.135653002618, ymax = 15.910337372231306; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((xmin, -0.23437499999999994*xmin + 2.9190625)--(xmax, -0.23437499999999994*xmax + 2.9190625), linewidth(2) + wrwrwr); /* line */ draw((-8.58,4.93)--(-10.46,-1.75), linewidth(2) + wrwrwr); draw((xmin, 0.049875311720698236*xmin-1.2283042394014965)--(xmax, 0.049875311720698236*xmax-1.2283042394014965), linewidth(2) + wrwrwr); /* line */ draw((0.38,2.83)--(-2.44,-1.35), linewidth(2) + wrwrwr); draw((-10.46,-1.75)--(0.38,2.83), linewidth(2) + wrwrwr); draw((-12.705589144106808,5.896934955650032)--(-10.46,-1.75), linewidth(2) + wrwrwr); draw((-8.58,4.93)--(-2.44,-1.35), linewidth(2) + wrwrwr); draw((-2.44,-1.35)--(-12.705589144106808,5.896934955650032), linewidth(2) + wrwrwr); /* dots and labels */ dot((-8.58,4.93),dotstyle); label("$A$", (-8.446284308486987,5.2411858348205556), NE * labelscalefactor); dot((0.38,2.83),dotstyle); label("B", (0.517043581954025,3.132167507657965), NE * labelscalefactor); dot((-10.46,-1.75),dotstyle); label("D", (-10.33819780785343,-1.4270338760611636), NE * labelscalefactor); dot((-2.44,-1.35),dotstyle); label("C", (-2.305319179395913,-1.0548541712677653), NE * labelscalefactor); dot((14.590544208361896,-0.5005962988348186),linewidth(4pt) + dotstyle); label("$E$", (14.721902314902065,-0.24846481088206912), NE * labelscalefactor); dot((-12.705589144106808,5.896934955650032),dotstyle); label("$F$", (-12.57127603661382,6.202650072203501), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] We take a point F on $AB$ such that $BF=BD$ now we know $BF=BD=AB+AC$ or that $AC=BF-AB=AF$ we have that by easy angle chase that $AD \bot FC$ so we can easily say $DF=DC$ by simple congruency. Now let $\angle ADB=2\alpha$ and so we have $\angle DAB=60-2\alpha$ so $\angle BFD=60+\alpha$ since $BF=BD$ and so $\angle CFD=30+\alpha$ and as $DF=DC$ we have $\angle FCD=30 +\alpha$ and so we get $\angle FCE =150-\alpha$ so $\angle FEC=\alpha$ and we are done $\blacksquare$.