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

2 comments:

  1. 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

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  2. <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

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