Sunday, August 14, 2011

Problem 657: Triangle, Incircle, Tangency Points, Parallel Lines

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

Click the figure below to see the complete problem 657.

Online Geometry Problem 657: Triangle, Incircle, Tangency Points, Parallel Lines.

Posted by Antonio Gutierrez at 5:34 PM
Labels: , , ,

4 comments:

  1. AjitAugust 15, 2011 at 3:48 PM

    See problem # 328. J & K are midpoints of AB & AC resply. Hence JK = BC/2 =(BE+CE)/2=(BD+CF)/2 = 4.5
    Ajit

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  2. Peter TranAugust 16, 2011 at 10:48 AM

    http://img143.imageshack.us/img143/1590/problem657.png
    Connect DF and AE
    1. We have (DEF)=(ADF) both angles look the same arc DF
    but (AHF) supplement to (DEF)
    So (AHF) supplement to ( ADF) >> A,D,F, H is cyclic
    Similarly we also have A,G,D,F cyclic
    5 points A,G,D,F,H are cyclic ( see picture)
    2. We have (EDF)= (FEC) both angles look the same arc EF
    And ( EDF)=(GHF) both angles look the same arc GF
    So ( FEC)=(GHF) and GH //BC

    3. Parallelogram AGEH have diagonals bisect each other so AJ/AB= ½ and KJ= 4.5
    Peter Tran

    ReplyDelete
  3. bttfSeptember 24, 2011 at 4:06 PM

    http://imgsrc.baidu.com/forum/pic/item/9c16fdfaaf51f3debc44053b94eef01f3b2979fe.jpg

    Chinese version

    ReplyDelete
  4. Sumith PeirisNovember 8, 2015 at 9:59 PM

    Let JK meet AE at L

    Since AGEH is a parellelogram AL = LE. But LK//EC so from mid point theorem LK = EC/2 = 2

    Similarly JL = BE /2 = 2.5

    So JK = 4.5

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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