Wednesday, January 22, 2014

Geometry Problem 964: Right Triangle, Cevians, Angles, 10, 20, 30 Degrees

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 964.

Online Geometry Problem 964: Right Triangle, Cevians, Angles, 10, 20, 30 Degrees

4 comments:

  1. 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!

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  2. 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°.

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  3. See http://fukuchandayo.web.fc2.com/GG_964.html

    Angle 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.

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