Monday, April 27, 2015

Geometry Problem 1113: Triangle, Perpendicular Bisector, 90 Degrees, Circle, Concyclic Points, Cyclic Quadrilateral

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

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Online Math: Geometry Problem 1113: Triangle, Perpendicular Bisector, 90 Degrees, Circle, Concyclic Points, Cyclic Quadrilateral.

Posted by Antonio Gutierrez at 1:21 PM
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2 comments:

  1. Jacob HA (EMK2000)April 27, 2015 at 7:17 PM

    By construction, O is the orthocenter of ΔCDE, thus CO⊥DE.

    ∠BOA = 2∠C
    ∠BDA = 180° − 2∠CDA = 2∠C (since DA=DC)
    Thus BDOA concyclic.

    Also, (since EB=EC)
    ∠BEA = 180° − 2∠C
    Thus BOAE concyclic.

    Hence, EAODB concyclic.

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  2. APOSTOLIS MANOLOUDISSeptember 11, 2016 at 12:37 AM

    Problem 1113
    Is <MDC=<MDA=<ODA=90-<ACB=<AEO so EAOD is cyclic.
    Is <OAB=90-<OAB/2=90-<ACB=<ODC so AODB is cyclic.
    Therefore E,A,O,D and B are concyclic.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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