Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Tr. BFE = Tr. BEA - Tr. BFA = Tr. BCA - Tr. BHA. Now draw a perpendicular HX to AB.It can be proven easily that HX =(2R²-25)/2R if R be the radius of the circle. Hence Tr. BFE = (R*R) - ((2R²-25)/2R)*(2R/2) = 25/2 = 12.5 sq. units.
<CBF=<ECF from tangent angle theorem, <FAB=<FCB from inscribed angles, and <FAB=<CEF from parallel property. Therefore <FCB=<FEC, making triangle CFB similar to triangle ECF. FB/CF=CF/FE, which results in CF^2=25=FB*FE=2*[FEB].