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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 964.
I do not have a full solution but a suggestion for others. Consider G the midpoint of BD. G is the circumcentre of Tr. ABD or GA=GD. For reasons that are still unclear to me, GAD is an equilateral triangle, In other words, AD=GA=GD or AD=BD/2 which in turn suggests that BAD is a 30°-60°-90° triangle. /_EDF is therefore=40° and thus x=(40+10)=50°.All that remains now is to prove that Tr. GAD is equilateral. Still beyond me though -- A lovely problem, Antonio!
Consider G pe BC astfel incatDG= ED=DC,rezulta EDG echilateral,EGB,GDC,BGD isoscele=>BG=EC=GD=ED=DC m<BDG=20°,m<BDE=40°.x=50°.
See http://fukuchandayo.web.fc2.com/GG_964.htmlAngle DEC=10 deg. ---> triangle DCE is isoceles ---> CD=DE.Put point G on BC, so that DC=DG ---> angle DGC=DCG=40 deg.---> angle CDG=100 deg. ---> angle GDE=60 deg.Triangle DGE is equilateral <--- DE(=CD)=DG.Angle GEB=180-70-60=50 deg.=angle GBE.---> GE=GB, and angle BGE is 80 deg.Points B, E, D are on circle G.---> angle BDE=angle BGE/2=40 deg.Angle BFE(=x)=40+10=50 deg.
Excellent solution FUKUCHAN