Monday, July 16, 2018

Geometry Problem 1365: Circle, Tangent, Secant, Chord, Parallel, Midpoint, Sketch, iPad Apps

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

Details: Click on the figure below.

Geometry Problem 1365: Circle, Tangent, Secant, Chord, Parallel, Midpoint, Sketch, iPad Apps.

4 comments:

  1. < AMB = < BFC < ACB = < AOB since ABOC is concyclic

    Hence ABMO is concyclic and so < AMO = < ABO = 90

    Therefore M is the midpoint of DE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. https://photos.app.goo.gl/BGn1ZyJuoHQCZUsQ7
    Connect OA, OB, OC
    Since AB and AC tangent to circle O => ∠ (BOA)= ∠ (COA)
    We have ∠ (BFC)= ½. ∠ (BOC)= ∠ (BOA)
    ∠ (BFC)= ∠ (BMA)= ∠ (BOA) => MOAB is cyclic quad.
    So ∠ (OMA)= ∠ (OBA) = 90 degrees => M is the midpoint of ED

    ReplyDelete
  3. hi , can you explain how you decide that MOAB is cyclic quad.
    thanks

    ReplyDelete
    Replies
    1. in quadrilateral MOAB , we have ∠ (BMA)= ∠ (BOA)
      so MOAB is cyclic

      Delete