Points$ D, E, F$ are chosen on the sides $AB$, $BC$, $AC$ of a triangle $ABC$, so that $DE = BE$ and $FE = CE$. Prove that the centre of the circle circumscribed around triangle $ADF$ lies on the bisectrix of angle $DEF$.
Problem
Source:
Tags: geometry
Rijul saini
25.12.2023 01:11
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.114cm); 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 = -10.56, xmax = 19.82, ymin = -4.59, ymax = 9.81; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-2.38,7.67)--(-5.06,-0.07)--(4.6,-0.31)--cycle, linewidth(0.7) + rvwvcq); /* draw figures */ draw((-2.38,7.67)--(-5.06,-0.07), linewidth(0.7) + rvwvcq); draw((-5.06,-0.07)--(4.6,-0.31), linewidth(0.7) + rvwvcq); draw((4.6,-0.31)--(-2.38,7.67), linewidth(0.7) + rvwvcq); draw(circle((-4.460754240611095,4.962141815403614), 3.414972058115072), linewidth(0.7) + dbwrru); draw(circle((-1.4901319210833297,3.2204263103371757), 3.4435693496523814), linewidth(0.7) + dtsfsf); /* dots and labels */ dot((-2.38,7.67),dotstyle); label("$A$", (-2.3,7.87), NE * labelscalefactor); dot((-5.06,-0.07),dotstyle); label("$B$", (-5.6,-0.49), NE * labelscalefactor); dot((4.6,-0.31),dotstyle); label("$C$", (4.96,-0.63), NE * labelscalefactor); dot((-2.242236187685546,-0.14000655434321624),dotstyle); label("$E$", (-2.16,0.05), NE * labelscalefactor); dot((-4.499971773810847,1.5473949517552392),dotstyle); label("$D$", (-5.12,1.23), NE * labelscalefactor); dot((-1.5000488765022317,6.663981380299111),dotstyle); label("$F$", (-1.26,6.97), NE * labelscalefactor); dot((-4.460754240611095,4.962141815403614),linewidth(4.pt) + dotstyle); label("$O$", (-4.7,5.27), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $O$ be the circumcentre of $\displaystyle \triangle ADF$. Then $\angle DOF = 2A$. On the other hand, $\angle DEB = 180^{\circ} - 2B$ and $\angle FEC = 180^{\circ} - 2C$, therefore $\angle DEF = 180^{\circ} - 2A$. Thus, $D,O,F,E$ are concyclic. $OD = OF$ implies that $O$ is the midpoint of the arc $DF$, and hence $EO$ bisects $\angle DEF$. $\square$