Wednesday, March 1, 2017

Geometry Problem 1320: Triangle, Incircle, Tangent, Chord, Circle, Parallel, Perpendicular, Collinearity

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

Details: Click on the figure below.

Geometry Problem 1320: Triangle, Incircle, Tangent, Chord, Circle, Parallel, Perpendicular, Collinearity.

4 comments:

  1. Extend HF at G , G on AB. Join G to K, meet DE at P. From 1315 DP=NE =>PKN isoceles, => GKH isoceles

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  2. Dear Antonio

    Can this be proved directly without using Pr 1315?

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  3. https://goo.gl/photos/NtvVeEVnYmjzHqrLA

    Let HF cut DB at L; LK and FA cut DE at N’ and G
    Per result of problem 1315 we have
    DG=MN
    GM=NE , GN=ME=DM
    triangle FGM similar to FAK so GM= KA.(FM/FK)
    Triangle LDN’ similar to LAK so DN’= KA.( LN’/LK)= KA.(FM/FK)
    So GM= DN’=NE => MN=MN’
    Due to symmetry of LN’K to HNK over axis of symmetry BK => H,N ,K are collinear

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