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Geometry ProblemLevel: Mathematics Education, High School, College.Click the figure below to see the complete problem 614.
Consider ∆BED and transversal CMFBy Menelau: (EM/MD) (BC/CE) (DF/FB) = 1Consider ∆CEA and transversal BHGBy Menelau: (EH/HA) (AG/GC) (CB/BE) = 1Note that (BC/CE) = (AD/EC) = (AG/GC) and (DF/FB) = (AD/BE) = (CB/BE) It follows (EM/MD) = (EH/HA)Hence HM ∥AD
Another way to solve using similar triangles.http://img687.imageshack.us/img687/6279/problem614.pngLet CF cut AD at H, BG cut AD at KLet a=BE, b=ECUsing similar triangles we haveBE/DK=GE/GD=b/(a+b) so DK=a/b*(a+b)HE/HA=HB/HK=BE/AK=a.b/(a+b)^2With the same way as above we have AH=b/a*(a+b)And EM/MD=EC/DH=a.b/(a+b)^2So EM/MD=HE/HA and HM //ADPeter 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 FBD (p), FCM(p’), FAH (p’’) concur in F; lines GCA (q), GBH (q’), GDM (q’’) concur in G then line BC (r), HM (r’) and line AD (r’’) concur or are parallel. Since AD//BC then also HM//AD//BC.
CH cuts AB at P1 and BM cuts CD at P2. Applying Ceva twice, once to M in triangle BCD, and once to H in BAC, we obtain1)EC/BE*(BF/FD=BE/AD)*DP2/CP2=12)BE/EC*(CG/AG=EC/AD)*AP1/BP1=1It follows that DP2/CP2=AD/EC and AP1/BP1=AD/BE.From Aubel II we know thatAH/HE=(AG/CG=AD/EC)+(AP1/BP1=AD/BE)DM/ME=(FD/BF=AD/BE)+(DP2/CP2=AD/EC)We have shown that AH/HE=DM/ME so we are done.