Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 706.
Monday, December 19, 2011
Problem 706: Triangle, Cevian, Circumcenters, Three Circumcircles, Concyclic Points
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∠EBO =∠OBA-∠EBA =(90°-∠BCA)-(90°-∠BDA)
ReplyDelete=∠BDA-∠BCA=∠DBC=∠EFO (since EF⊥DB and OF⊥CB)
∴ E,O,F,B are concyclic
Note that OE and OF are perpendicular bisectors of AB and AC.
ReplyDeleteSo angle(EOF) supplement to angle(EOF) and E,O,F,B are concyclic