Geometry Problem

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

Click the figure below to see the complete problem 964.

## Wednesday, January 22, 2014

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

Labels:
10,
20,
30 degrees,
angle,
cevian,
right triangle

Subscribe to:
Post Comments (Atom)

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°.

ReplyDeleteAll 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°.

ReplyDeleteSee http://fukuchandayo.web.fc2.com/GG_964.html

ReplyDeleteAngle 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

ReplyDelete