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 677.
http://img542.imageshack.us/img542/4607/problem677.pngExtend FE to cut BC at H ( see picture)Let AD=BC=a and DF=2aTriangle EMC similar to tri. EDA ( case AA) So EM/MD=MC/AD=1/2Triangle EMH similar to tri. EDF ( case AA) So MH/DF=ME/ED=1/2 and MH=a , CH=1.5aTriangle GCH similar to tri. GDF so CH/DF=GC/GD=1.5a/2a = 3/x x=4Peter Tran
Draw DH parallel FE (E on AC). Thenx/3 = HE/EC = (HE/AE)(AE/EC) = (DF/AF)(AD/MC)= (2/3)(2)= 4/3x = 4
Problem 677Are triangles MEC and AED similar then CE/EA=MC/AD=1/2. In triangle ACD apply the of Menelaus’ theorem intersected by EGF. CE/EA.AF/FD.DG/GC=1 or 1/2.3/2.x/3=1 or x=4.STOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE