Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, College.Click the figure below to see the complete problem 616.
http://img827.imageshack.us/img827/9192/problem616.pngLet BG cut CD at L and CF cut AB at MLet a=BE , b=EC and c=AB=CDTriangle(EGC) similar to Triangle (DGA); Triangle(BGA) similar to Triangle (LGC) case AASo GC/GA=b/(a+b)=CL/AB and CL=b.c/(a+b)With the same way as above BM=a.c/(a+b)Triangle (BHM) similar to Triangle (LHC) case AAHB/HL=BM/CL=a/bSo HB/HL= BE/ECAnd HE//CD ( Thales theorem on parallel lines)Peter Tran
There is no need for E to be on BC.The dual of Pappus hexagon theorem says that if lines p, p’, p’’ concur, and lines q, q’, q’’ concur, and if you call (I'll call a^b the intersection point of lines a and b)r the line joining p^q’ with p’^q,r’ the line joining p’^q’’ with p’’^q’,r’’ the line joining p^q’’ with p’’^q,then lines r, r’, r’’ concur or are parallel.In this problem lines FAE (p), FDB (p’), FHC (p’’) concur in F; lines GHB (q), GAC (q’), GDE (q’’) concur in G then line AB (r), line DC (r’) and line HE (r’’) concur or are parallel. Since AB//DC then also HE//AB//DC.
CF cuts AB at P1 and BG cuts CD at P2. Let O be the center of the parallelogram. By Ceva in triangle ABC with point F and in triangle BCD with point G, we obtain BE/EC*(CO/AO=1)*AP1/BP1=1EC/BE*(BO/DO=1)*DP2/CP2=1Therefore AP1/BP1=EC/BE and DP2/CP2=BE/EC. Because AB=CD, BH/HP2=BP1/CP2=BE/CE so EH is parallel to CD.