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 667.
http://img716.imageshack.us/img716/570/problem667.pngPer the result of problem 666, quadrilateral CEDB is cyclicSo (ECD)=(EBD)= β = (EDF)And (CBF)=(CDE)= αIn triangle EGD we have (CGD)=(GDE)+(GED)= β+ ( 180- α- β)=180- αSo (CGF) supplement to (CBF) and quadrilateral CGFB is cyclicWe have EF.EB=EG.EC= ED^2And x^2=4 x 9= 36 ; x=6Peter Tran
Let EB intersect circle (O) again at H.Join CB, AB∠BCG = ∠BCA + ∠ACG = ∠BCA + ∠CBA = ∠BAD = ∠BFD = ∠GFESo G, C, B, F are concyclicHence x² = EF.EB = EG.EC = 4 x 9 = 36∴ x = 6
< ECD = < ABC hence < ECB = < BAD = < BFD implying that BCGF is con cyclic So EG. EC = EF. EB = ED^ 2 from whichx^2 = 4(4+5) and so x = 6Sumith PeirisMoratuwaSri Lanka