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

4 comments:

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

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  3. 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.

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  4. 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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