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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1244.
https://goo.gl/photos/FGTGW9iSkgxRKfx16Connect BF, BC and BDNote that C is the midpoint of arc BGTriangle BEC similar to FBC ( case AA)So CB^2=CE.CFTriangle DCB similar to BCA ( case AA)So CB^2=CD.CAAnd CE.CF=CD.CA or 4x 10= 3(x+3)So x= 31/3
Let AB cut the circle at G. In isoceles Tr. BCG, CG^2 - CE^2 = BE.GE = CE.EF = 24So CG^2 = 24 + 16 = 40Now < CGB = CBG = < GDA Hence < CGD = < DGB - < CGB = < DGB - < GDA = < GADHence CG^2 = CD.CASo 40 = 3(3+x) from whence x = 31/3Sumith PeirisMoratuwaSri Lanka
2nd solution < CFD = < CGD = < EAD as before Hence AFED is cyclic and the result is easily calculated Sumith PeirisMoratuwaSri Lanka