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Geometry ProblemPost your solution in the comment box belowLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 918.
Note that triangles AFI, AIB and IDB are similar ..( case AA)so we have BD/ID=IF/AF => ID^2=BD. AF= 36DF=2.ID=12
Extend CB to E such that DE = FA = 9 so ACE is an isoceles Tr. similar to isoceles Tr. CFD< IDC = 90 - C/2 = A/2 + B/2 and so since < IBD = B/2, < BID = A/2 = IEDHence DI is tangential to circle IBE and so (x/2)^2 = 4 X 9 = 36 from which x = 12Sumith PeirisMoratuwaSri Lanka