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 diagram below to enlarge it.
http://s17.postimg.org/5cijbp9e7/Pro_1133.pngDraw line BMN// GH ( M on EC and N on GC)Since GH tangent to circle EFC so ∠ (EDF)= ∠ (GEF)= ∠ (BEH)=∠ (MBE)Triangle FED similar to MEB ( case AA)So EM/EF= EB/ED ……..(1)Triangle AED similar to CEB ( case AA)So EC/EA=EB/ED…… (2)From (1) and (2) we have EM/EF=EC/EA => MF//CAAnd FA/FB= MN/MB …..(3)Since NB//GH => EG/EH=MN/MB…..(4)From (3) and (4) we have FA/FB= EG/EH
<BEH=<GEA=<FDE, so <CHE=<CBE-<BEH=<ADE-<FDE=<AFD ( 1 ).Let the circle (CGH) intersect CE at P and, easily, triangles GPH and ADB are similar and <GPE=<EHC(=<ADF as per (1)), consequently DF and PE are homologous lines within the 2 similar triangles, thus solving our problem.