Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to enlarge the problem 945.
http://img15.imageshack.us/img15/4941/dfl0.pngLet B’, F’, D’ are the projections of B, F, D over ACApply Ceva’s theorem on triangle BFD with Cevians BG, FC and DA(GF/GD).(CD/CB).(AB/AF)= 1 => (GF/GD)=(CB.AF)/(CD.AB) ……. (1)Use properties of similar triangles CDD’ to CBB’ and triangles AFF’ to ABB’ we haveFF’/DD’= (FF’/BB’)/(DD’/BB’)=(AF/AB)/(CD/CB)………(2)We also have GF/GD= HF’/HD’… ( Thales theorem on parallel lines)From (1) and ( 2) we have GF/GD=FF’/DD’=HF’/HD’ => triangles FF’H similar to DD’H … ( case SAS)So GF/GD=HF/HD => HG is angle bisector of angle ( FHD)
if X=intersection of lines DF and AC then (A,C,E,X)=1 angle GHX is 90 so H lies on Apollonius circle with diameter GX so FG/GD=FH/HD by the angle bisector theorem we have what we wanted