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

Labels:
collinear,
Menelaus' theorem,
parallel,
parallelogram

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http://img94.imageshack.us/img94/1274/problem659.png

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.