Monday, July 21, 2008

Elearn Geometry Problem 137

Geometry Problem 137 Concyclic points

See complete Problem 137
Orthic Triangle, Altitudes, Perpendicular, Incircle, Collinear Points, Parallelogram. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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1 comment:

  1. c .t . e. oMarch 10, 2010 at 2:57 PM

    DM = GK (from P135) =>
    GM = DK & G,N,H,M,K,D in a circle ( P 136)
    =>
    MK//GD (1) ( GMDK trapezoid cyclic)
    =>
    KM perpen BE (2)
    ML perpen BE (3) ( EHLM kite )
    (2) & (3) =>

    K,M,L colline (4)
    =>
    KL//GD (5)

    NGE = NDA ( have same arc NH )
    A + ABE = 90°
    BDN + NDA = A + NDA = A + NGE ( BDN = A P134 )
    =>
    NGE = ABE =>
    NG//AB (6)

    LN //AB (7) ( LN perp FC, FLHN kite)
    (6) & (7) =>

    G,N,L collin

    2)
    LK//GD//AC (8) ( from (5) & GD//AC )
    LG//AB//KD (9) ( from (6) & (7) & DK//AB)

    3)
    LGDK paralelogr (10) ( from (8)&(9) )

    GL = GM = DK ( from (10) & (1)
    -------------------------------------------

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