Geometry Problem
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Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 918.
Thursday, September 5, 2013
Problem 918: Triangle, Incenter, Incircle, Angle Bisector, Perpendicular, Metric Relations
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Note that triangles AFI, AIB and IDB are similar ..( case AA)
ReplyDeleteso we have BD/ID=IF/AF => ID^2=BD. AF= 36
DF=2.ID=12
Extend CB to E such that DE = FA = 9 so ACE is an isoceles Tr. similar to isoceles Tr. CFD
ReplyDelete< IDC = 90 - C/2 = A/2 + B/2 and so since < IBD = B/2, < BID = A/2 = IED
Hence DI is tangential to circle IBE and so (x/2)^2 = 4 X 9 = 36 from which x = 12
Sumith Peiris
Moratuwa
Sri Lanka