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