Monday, December 19, 2011

Problem 706: Triangle, Cevian, Circumcenters, Three Circumcircles, Concyclic Points

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 706.

Online Geometry Problem 706: Triangle, Cevian, Three Circumcenters, Circumcircles, Concyclic Points<br />

5 comments:

  1. ∠EBO =∠OBA-∠EBA =(90°-∠BCA)-(90°-∠BDA)
    =∠BDA-∠BCA=∠DBC=∠EFO (since EF⊥DB and OF⊥CB)
    ∴ E,O,F,B are concyclic

    ReplyDelete
  2. Note that OE and OF are perpendicular bisectors of AB and AC.
    So angle(EOF) supplement to angle(EOF) and E,O,F,B are concyclic

    ReplyDelete
  3. Replies
    1. To Sumith

      You 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

      Delete
  4. EO perpend AB, OF perp BC =>BE'E + BF'O = 180
    (E' on AB, F' on BC)

    ReplyDelete