Sunday, April 29, 2012

Problem 745: Circle, 90 Degrees, Chord, Congruence, Perpendicular

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

Click the figure below to see the problem 745 details.

Online Geometry Problem 745: Circle, 90 Degrees, Chord, Congruence, Perpendicular, Metric Relations.

Posted by Antonio Gutierrez at 1:56 PM
Labels: , , , ,

9 comments:

  1. UnknownApril 30, 2012 at 5:39 AM

    The slution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJNkJFWExzWE1NSW8

    ReplyDelete
  2. UnknownApril 30, 2012 at 5:50 AM

    The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJM0w2aWEyS3lMNEk

    ReplyDelete
  3. Peter TranApril 30, 2012 at 8:49 AM

    http://img715.imageshack.us/img715/7306/problem745.png

    Since ∠ACD=90 => AD is a diameter
    Since AB=BD => BO perpendicular to AD
    Let BE cut AD at F and r = radius of the circle
    Since EF//CD => AE/AC=AF/AD => AF=10/7 .r
    Note that ∠EAF=∠OBF= alpha
    In ∆ BOF , tan(alpha)=OF/OB=3/7
    In ∆CAD, tan(alpha)=x/7=3/7
    So x=3

    ReplyDelete
  4. AjitMay 2, 2012 at 4:33 AM

    B is the midpoint of arc AD and BE is perpendicular to chord AC. Hence by the Broken Chord Theorem, 5 = 2 + x or x = 3
    If you wish to prove the theorem all you need to do is extend AC to F such that CF =CD=x. Now /_ABD = /_ACD = 2*/_AFD and B being the midpoint of arc AD lies on the perpendicular bisector of AD. Therefore B is the circumcentre of triangle ADF and BE is perpendicular to AF so AE = EF or 5 = 2 + x or x = 3
    Ajit

    ReplyDelete
  5. Eder Contreras OrdenesJuly 4, 2012 at 2:37 PM

    http://img18.imageshack.us/img18/6838/p745resuelto.png

    (look at the picture)

    Note that ABCD is cyclic, then 7*BD=BD*x+2sqrt(2)*BD*sqrt(2)
    <=>x=3

    Greetings

    ReplyDelete
  6. Sumith PeirisSeptember 6, 2015 at 12:56 PM

    AD is a diameter so < BDA = 45 = < BCA. Hence BE = EC = 2 and so applying Pythagoras to Tr. ABE, AB^2 = 29 from which we have AD^2 = 58.
    So x^2 = 58 - 7^2 = 9 and thus x = 3

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisSeptember 6, 2015 at 1:06 PM

    Another proof without using Pythagoras...

    Extend EC to F such that CF = 3. So Tr. ABF is isoceles and hence B is the centre of circle ADF. Hence < AFD = 1/2 of < ABD = 45 and so x = CF = 3

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  8. Sumith PeirisSeptember 6, 2015 at 1:17 PM

    Still another method much simpler than both previous proofs...,

    Find G on AE such that GE = 2. Then prove that Tr.s AGB & BDC are congruent SAA and the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  9. APOSTOLIS MANOLOUDISJuly 1, 2016 at 5:25 AM

    Problem 745
    From the theorem of broken chord (Archimedes) is arc AB=arc BD then AE=EC+CD.
    So 5=2+x .Therefore x=3
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete

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