An acute triangle ABC is inscribed in a circle C. Tangents in A and C to circle C intersect at F. Segment bisector of AB intersects the line BC at E. Show that the lines FE and AB are parallel.
Problem
Source: Saudi Arabia JBMO training test 2 2019, P2
Tags: geometry
Frestho
20.04.2020 04:42
[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 1; /* changes label-to-point distance */ pen dps = linewidth(0.05) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.257069257943465, xmax = 8.077596664374909, ymin = -7.082954858162459, ymax = 5.167662362298842; /* image dimensions */ pen wrwrwr = rgb(0, 0, 0); /* draw figures */ draw(circle((-3.28,0.49), 3.5392654605157836), linewidth(1) + wrwrwr); draw((-6.2,-1.51)--(-0.45166414020892987,-1.6376551093210365), linewidth(1) + wrwrwr); draw((-4.953576852857056,3.6085798879587925)--(-6.2,-1.51), linewidth(1) + wrwrwr); draw((-4.953576852857056,3.6085798879587925)--(-0.45166414020892987,-1.6376551093210365), linewidth(1) + wrwrwr); draw((xmin, -1.46*xmin-10.562)--(xmax, -1.46*xmax-10.562), linewidth(1) + wrwrwr); /* line */ draw((xmin, 1.329320643839513*xmin-1.0372486436592818)--(xmax, 1.329320643839513*xmax-1.0372486436592818), linewidth(1) + wrwrwr); /* line */ draw((xmin, -0.2435095621102043*xmin-0.30871136372147)--(xmax, -0.2435095621102043*xmax-0.30871136372147), linewidth(1) + wrwrwr); /* line */ draw((-2.0126199649314187,0.18138084263323448)--(-3.4147208487403775,-5.576507560839048), linewidth(1) + wrwrwr); draw(circle((-3.347360424370189,-2.5432537804195245), 3.0340016353325634), linewidth(1) + linetype("2 2") + wrwrwr); draw((-6.2,-1.51)--(-2.0126199649314187,0.18138084263323448), linewidth(1) + wrwrwr); /* dots and labels */ dot((-3.28,0.49),dotstyle); dot((-3.199311608803878,0.4667167200975916), linewidth(4pt)+dotstyle); dot("$O$", (-3.199311608803878,0.7067167200975916), NW * labelscalefactor); dot((-6.2,-1.51),dotstyle); label("$B$", (-6.120438450651395,-1.2859899627131193), SW * labelscalefactor); dot((-0.45166414020892987,-1.6376551093210365),dotstyle); label("$C$", (-0.3687623434477571,-1.4218563274502132), SE * labelscalefactor); dot((-4.953576852857056,3.6085798879587925),dotstyle); label("$A$", (-4.852352379771853,3.8316431090507517), NW * labelscalefactor); dot((-3.4147208487403775,-5.576507560839048),linewidth(4pt) + dotstyle); label("F", (-3.335177973540972,-5.3846252989487855), NE * labelscalefactor); dot((-2.0126199649314187,0.18138084263323448),linewidth(4pt) + dotstyle); label("E", (-1.931225537924336,0.3670508082548568), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Since $E$ lies on the perpendicular bisector of $\overline{AB}$, triangle $EAB$ is isosceles. Therefore, $\angle BEC = \angle BAE + \angle EBA = 2\angle A$. Since $\angle BFC = 180^{\circ}-2\angle A$, $E$ lies on $(BCF)$. It is well known that $O$ lies on this circle too and $\overline{OF}$ is a diameter of this circle, so $\angle OEF = 90^{\circ}$. Because $\overline{FE}$ and $\overline{AB}$ are perpendicular to the same line, they must be parallel.