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 702.
Join G to F. /_EDF=/_EGF & /_DGE=/_DFE so /_DGF=/_DGE+/_EGF=/_EDF+/_DFE=/_DEB=/_ACB. Hence etc.
http://img809.imageshack.us/img809/2501/problem702.pngConnect GF ( see picture)Since quadrilateral GDEF cyclic so m(GDE)=m(GFC)Note that m(GDE) supplement to angle(GAC) ( DE//AC)So angle(GFC) supplement to angle (GAC) And A,G,E,F concyclicPeter Tran
DE//AC ---> angle (BED)=(BCA)cyclic quadrilateral DEFG ---> (BED)=(BGF)Therefore: angle(BGF)=(BCA).Then the quadrilateral AGFC is cyclic.
Join GFAngle DGF = angle BED = angle BCASo A,G,F,C are concyclic
Triangle BGF ~ Triangle BED ~ Triangle BCABG:BC = BF:BABG.BA = BF.BCA, G, F, C are concyclic