Monday, October 17, 2011

Problem 678: Triangle, Simson Line, Circumcircle, Tangent, Parallel, Perpendicular, Collinear Points

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

Click the figure below to see the complete problem 678.

Online Geometry Problem 677: Parallelogram, Midpoint, Diagonal, Metric Relations
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3 comments:

  1. Peter TranOctober 17, 2011 at 10:16 PM

    http://img818.imageshack.us/img818/2386/problem678.png
    Connect BE ( see picture)
    Note that quadrilateral BGED is cyclic
    So ( BGD)=(BED)
    But (BED)=(BAM) both angles face the same arc
    So (BAM)=(BGD) and AM//GDF
    Peter Tran

    ReplyDelete
  2. PravinOctober 18, 2011 at 3:59 AM

    <MAG
    = <MAB (same angle)
    = <ACB (angle in the alternate segment)
    = <AEB (angles in the same segment)
    = <DEB (same angle)
    = <DGB (B,G,E,D are concyclic; angles in the
    same segment)
    = <DGA (same angle)
    So AM // GD

    ReplyDelete
  3. Sailendra TSeptember 29, 2017 at 10:43 AM

    considering usual notations, let A,B and C be the angles of the triangle ABC
    hence m(MAB)=C (alternate-segment theorem)
    Therefore m(MAC) = A+C ---------(1)
    Since ADC is right triangle, m(DAC) = 90-C => m(BAD)=A-(90-C) = A+C-90
    Observe that A,G,E and F are concyclic
    => m(BAD)=m(GAE)=m(GFE)=A+C-90
    => m(GFC)=A+C -----------(2)
    From (1) and (2) GDF//AM

    ReplyDelete

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