Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 624.
Thursday, June 16, 2011
Problem 624: Triangle, Incenter, Excenter, Incircle, Excircle, Similarity
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∠EAC = ∠BAI (∵ each = A/2)
ReplyDelete∠AEC
=∠IEC
=∠IBC (∵B,I,C,E are concyclic)
=B/2
=∠ABI
Hence ∆AIB ~ ∆ACE
If AC meets the excircle at X < ECX = 1/2 BCX = A/2 + B/2. So < AEC = < B/2
ReplyDeleteSo Tr. s ABI and AEC both have angles < A/2 and B/2 and are therefore similar
Sumith Peiris
Moratuwa
Sri Lanka
Or since IBEC is con cyclic the result is obvious
ReplyDelete