Friday, September 9, 2011

Problem 667: Intersecting Circles, Secant, Tangent, Metric Relations, Mind Map

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

Click the figure below to see the complete problem 667.

Problem 667: Intersecting Circles, Secant, Tangent, Metric Relations, Mind Map.

Posted by Antonio Gutierrez at 4:03 PM
Labels: , , , ,

3 comments:

  1. Peter TranSeptember 9, 2011 at 9:53 PM

    http://img716.imageshack.us/img716/570/problem667.png
    Per the result of problem 666, quadrilateral CEDB is cyclic
    So (ECD)=(EBD)= β = (EDF)
    And (CBF)=(CDE)= α
    In triangle EGD we have (CGD)=(GDE)+(GED)
    = β+ ( 180- α- β)=180- α
    So (CGF) supplement to (CBF) and quadrilateral CGFB is cyclic
    We have EF.EB=EG.EC= ED^2
    And x^2=4 x 9= 36 ; x=6
    Peter Tran

    ReplyDelete
  2. PravinSeptember 9, 2011 at 10:06 PM

    Let EB intersect circle (O) again at H.
    Join CB, AB
    ∠BCG
    = ∠BCA + ∠ACG
    = ∠BCA + ∠CBA
    = ∠BAD
    = ∠BFD
    = ∠GFE
    So G, C, B, F are concyclic
    Hence x² = EF.EB = EG.EC = 4 x 9 = 36
    ∴ x = 6

    ReplyDelete
  3. Sumith PeirisFebruary 4, 2016 at 9:17 AM

    < ECD = < ABC hence < ECB = < BAD = < BFD implying that BCGF is con cyclic

    So EG. EC = EF. EB = ED^ 2 from which

    x^2 = 4(4+5) and so x = 6

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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