Thursday, September 5, 2013

Problem 918: Triangle, Incenter, Incircle, Angle Bisector, Perpendicular, Metric Relations

Geometry Problem
Post your solution in the comment box below

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 918.

Online Geometry Problem 918: Triangle, Incenter, Incircle, Angle Bisector, Perpendicular, Metric Relations.

2 comments:

  1. Note that triangles AFI, AIB and IDB are similar ..( case AA)
    so we have BD/ID=IF/AF => ID^2=BD. AF= 36
    DF=2.ID=12

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  2. 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 = 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

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