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

Labels:
angle bisector,
incenter,
incircle,
metric relations,
perpendicular,
triangle

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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