Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 659.
Sunday, August 21, 2011
Problem 659: Parallelogram, Parallel Lines, Collinear Points
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ReplyDeleteDraw EL //BC ,EL cut CH at L ( see picture)
BCLE is a parallelogram
Triangle CGF similar to triangle EGA ( case AA)
So GC/GE=GF/GA or CG/CE=FG/FA
But CG/CE= GH/EL = GH/AD ( triangle CGH similar to CEL)
So GH/AD=FG/FA and triangle FGH similar to tri. FAD ( case SAS)
Angle( GFH)=angle (AFD)
So D,H,F are collinear
Peter Tran
Peter, do you have mail? I have nice problems so that we can discuss them!
ReplyDeleteSlice
ReplyDeleteYou can email to me at vstran@yahoo.com
Peter Tran
Call K the point where line CF meets line DB and O the centre of ABCD.
ReplyDeleteTriangles OCK, OAE are congruent (OC=OA and the adjacent angles) so OE=KO.
AECK is a parallelogram, having diagonals that cut each other in their middle point O, so KA//CG.
Triangles ABK and GHC have parallel sides so, by Desargues' affine theorem, are perspective, that is lines KC, AG, BH concur, that is line BH passes through F.