Saturday, July 30, 2016

Geometry Problem 1240: Triangle, Circle, Diameter, Circumcircle, Circumcenter, Perpendicular. Mobile Apps

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1240.


Geometry Problem 1240: Triangle, Circle, Diameter, Circumcircle, Circumcenter, Perpendicular. Mobile Apps

4 comments:

  1. If < DAC = A then < DEB = A and < DOB = 2A

    Now DOB being isoceles < DBO = 90-A
    But < DCA = 90-A

    It follows that DBOC is concyclic and < CDA being 90, < BOC is also = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Correction to last line.......

      If BO meets AC at X, it follows that DBXC is concyclic and < CDA being 90, < BXC is also = 90

      Delete
  2. https://goo.gl/photos/LyXa4bZgSQcpQGuR9

    Draw diameter BF of circle O , lines AE and CD
    Since AD⊥DC= > DC will cut circle O at F
    Since AE⊥CE => AE will cut circle O at F
    In triangle ABF , C is the orthocenter so AC⊥BO

    ReplyDelete
  3. let angle CAB = A, ABC = B and ACB = C

    => angle DEC = DEB = DAC = A => triangles ABC & EBD are similar

    Extend BO to meet the circle at F and join FE
    => <FED = 90+A

    Since FEDB is cyclic quadrilateral <DBF = <DBO = <ABO = 90-A

    So if a triangle is formed by extending AC to intersect BO at a point P - it is right angle at P as <PAB = A & PBA = 90-A

    ReplyDelete