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


  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!

  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°.

  3. See

    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.

  4. Alternatively one could use trigonometry. Angle DEC = 10 deg, so CD=DE. Let CD=1 (unit length). AD = DE*cos(20 deg), BA=AC*tan(40 deg). DE=1, AC=1+AD=1+cos(20 deg)
    Tan(angle ABD) = AD/BA=cos(20deg)/(1+cos(20 deg))/tan(40 deg) = tan(30)
    Angle ABD = 80 - X = 30. So X = 50 deg.