Friday, April 19, 2013

Problem 871: Brahmagupta's Theorem, Cyclic Quadrilateral, Perpendicular Diagonals, Midpoint

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

Click the figure below to see the complete problem 871.

Online Geometry Problem 871: Brahmagupta's Theorem, Cyclic Quadrilateral, Perpendicular Diagonals, Midpoint

Posted by Antonio Gutierrez at 4:43 PM
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4 comments:

  1. nakaApril 20, 2013 at 2:15 AM

    Angle MCE = Angle EDF = Angle MED - 90 = Angle MEC
    So triangle MCE is isosceles and so as triangle BEM for symmetry.
    Hence BM = EM = CM
    Q.E.D.

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  2. Bogdan RusuApril 22, 2013 at 9:19 PM

    My solution http://www.docstoc.com/docs/153902983/Problem-871

    ReplyDelete
  3. EditorApril 25, 2013 at 7:48 AM

    Far more general:

    If ABCD is cyclic quadrilateral and its diagonals cuts in E, then orthocenter of tr ABE, circumcenter of tr DEC and E itself are colinear.

    In this particular case, EF contains the orthocenter of tr AED so must contain the circumcenter of tr BEC, but this is M.
    Done.

    ReplyDelete
  4. APOSTOLIS MANOLOUDISSeptember 1, 2016 at 12:11 PM

    Problem 871
    Let x=<BCA=<BDA=<AEF (perpendicular line) or <MEC=x.Then ME=MC.
    But <EBC=90-x=<BEM.So BM=EM.Therefore BM=MC.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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