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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 834.
Note that triangle BFA similar to triangle GFC ( case AA)so BF/FG=FA/FCTriangle BFC similar to triangle EFA ( case AA)so FE/BF=FA/FC => BF/FG=FE/BF => BF=GM(FG, FE)
From ΔAEF~ΔCBF, AF/FC = FE/BFFrom ΔABF~ΔCGF, AF/FC = BF/FGHence, BF/FG = FE/BFBF^2 = FG×FEBF = GM(FG,FE) = √(FG×FE)
Daca notam cu P intersectia ,paralelei prin D la BG ,vom obtine: paralelogramul BFDP, AF = PC=>AP=FC.In continuare aplicam teorema fundamentala a asemanarii in triunghiurile CFG,PD||FG si respectiv APD,EF||PD =>PD/FG=CP/CF=AF/CF=AF/AP=FE/PD=>PD=GM(FG,FE),qed.
triangle CGB similar to triangle ABE so CG/AB=CB/AE. But CG/AB=FG/BF, and CB/AE=BF/FE, so substituting these values in, we get FG/BF=BF/FE, or BF^2=FG*FE
EX is parallel to CD such that X is on AC. Then BF/FG=AB/CG=AE/BC=AE/AD=EX/CD=EX/AB=FE/BF, second equality due to fact that triangles ABE and CGB are similar. Result follows from BF/FG=FE/BF.