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.

Posted by Antonio Gutierrez at 5:39 AM
Labels: , , , , , , , ,

6 comments:

  1. c.t.e.oMarch 1, 2017 at 8:47 AM

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

    ReplyDelete
  2. Sumith PeirisMarch 3, 2017 at 10:52 PM

    Dear Antonio

    Can this be proved directly without using Pr 1315?

    ReplyDelete
  3. Peter TranMarch 10, 2017 at 11:54 PM

    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

    ReplyDelete
  4. takahiro4March 29, 2017 at 11:34 AM

    Angle FHN=HNE=MNK, so they are colinear.

    ReplyDelete
  5. Peter TranMarch 29, 2017 at 7:59 PM

    Please explain the reason of " HNE=MNK" in your solution
    Peter Tran

    ReplyDelete

Subscribe to: Post Comments (Atom)