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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 624.
∠EAC = ∠BAI (∵ each = A/2)∠AEC =∠IEC =∠IBC (∵B,I,C,E are concyclic)=B/2=∠ABIHence ∆AIB ~ ∆ACE
If AC meets the excircle at X < ECX = 1/2 BCX = A/2 + B/2. So < AEC = < B/2 So Tr. s ABI and AEC both have angles < A/2 and B/2 and are therefore similar Sumith PeirisMoratuwaSri Lanka
Or since IBEC is con cyclic the result is obvious