Saturday, December 17, 2011

Problem 704: Triangle, Orthocenter, Altitude, Midpoint, Perpendicular

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

Click the figure below to see the complete problem 704.

Online Geometry Problem 704: Triangle, Orthocenter, Altitude, Midpoint, Perpendicular

Posted by Antonio Gutierrez at 6:21 PM
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4 comments:

  1. Peter TranDecember 17, 2011 at 8:52 PM

    Quadrilateral BEHD is concylic . BH is the diameter of circle (EBHD).
    Quadrilateral AEDC is concylic . AC is the diameter of circle (AEDC). These 2 circles intersect at E and D so line connected 2 centers will perpendicular to ED.
    Peter Tran

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  2. AnonymousDecember 17, 2011 at 11:34 PM

    AEDC is a cyclic quadrilateral, AC is the diameter and F is the circumcenter:then EF = DF.
    EBDH is a cyclic quadrilateral,BH is the diameter and M is the circumcenter:then EM = DM.
    Therefore EMDF is a Kite, then FM is perpendicular to DE.

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  3. Sumith PeirisAugust 8, 2015 at 10:31 PM

    Tr.s FEM & FDM are congruent SSS, hence FM is the perpendicular bisector of DE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. APOSTOLIS MANOLOUDISJune 28, 2016 at 3:35 AM

    Problem 704
    Is ME=MD=BH/2, FE=FD=AC/2.Then the points F, M belong to the perpendicular bisector of ED.Therefore FM is perpendicular to DE.
    MANOLOUDIS APOSTOLIS FROM GREECE

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