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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to enlarge the problem 947.
Angle ABE = angle AFE = angle GBDSo BDFG concylic Angle FAC = angle FBG = angle FDG QED
Let AF and BC intersect at P. Since ABFC concyclic, (AD+DP)×FP = (CG+GP)×BP ... (1)Since BDGF concyclic, DP×FP = GP×BP ... (2)(1) - (2): AD×FP = CG×BP ... (3)(2) / (3): DP/AD = GP/CGHence, DG//AC.
BDGFconcyclic=>< BFD= DG||AC (formeazacu secanta BC unghiuri corespondente congruente)
BFD=BGD,BFA=BCA, BGD=BCA,=> DG||AC