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

Labels:
cevian,
circle,
circumcenter,
circumcircle,
concyclic,
cyclic quadrilateral,
triangle

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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

Sorry Peter cannot understand

ReplyDeleteTo Sumith

DeleteYou are right. " OE and OF are perpendicular bisectors of AB and AC." is not enough to conclude that E,O,F,B are cocyclic.

addition work to show that angle(ABE)= angle(CBF) is required.

Thanks Sumith

EO perpend AB, OF perp BC =>BE'E + BF'O = 180

ReplyDelete(E' on AB, F' on BC)