Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 659.
http://img94.imageshack.us/img94/1274/problem659.pngDraw EL //BC ,EL cut CH at L ( see picture)BCLE is a parallelogramTriangle CGF similar to triangle EGA ( case AA)So GC/GE=GF/GA or CG/CE=FG/FABut 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!
SliceYou can email to me at firstname.lastname@example.orgPeter Tran
Call K the point where line CF meets line DB and O the centre of ABCD.Triangles 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.