Saturday, November 14, 2009

Problem 387. Triangle, Angle bisector, Perpendicular, Concyclic points

Proposed Problem
Click the figure below to see the complete problem 387.

Problem 387. Triangle, Angle bisector, Perpendicular, Concyclic points.
See also:
Complete Problem 387
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:36 PM
Labels: , , ,

9 comments:

  1. AnonymousNovember 14, 2009 at 10:45 PM

    For treangles BFH, BFE, BFG, DFD Phifagor theorem BF^2=BH^2+FH^2 and etc. Then BF is diametr of this circle.

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  2. vijay9290009015November 18, 2009 at 10:09 PM

    join B,F.sinve angleBHF+angleBGF=180 so B,H,F,S lie on same circle.simillarly the points B,E,F,D also concyclic.and angle BHF=BEF=90 so the points B,H,E,F are concylic and the points B,G,D,F are concyclic hence the points B,H,E,F,D,G discribe the same circle

    ReplyDelete
  3. Sumith PeirisAugust 26, 2015 at 2:53 AM

    There are 4 right angles at E, F, G and H and all these lie on a circle with BF as diameter

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Sumith PeirisSeptember 16, 2023 at 12:56 AM

      Dear Antonio - AF and CF need not be angle bisectors for this result to hold. Any point F inside the Triangle ABC would just as do

      Delete
    2. AnonymousSeptember 17, 2023 at 5:20 PM

      Dear Sumith
      You are correct; the condition of bisectors is not necessary. Perhaps I was thinking about other discoveries that have yet to be explored. Thank you.
      Antonio

      Delete
    3. AnonymousSeptember 17, 2023 at 5:38 PM

      Dear Sumith,
      You are correct; the condition of bisectors is not necessary. Perhaps I was thinking about other discoveries that have yet to be explored. Thank you.
      Antonio

      Delete
    4. Sumith PeirisSeptember 17, 2023 at 7:39 PM

      Thanks Antonio. I should have said "There are 4 right angles at D,E,H,G and all these points lie on the circle with BF as Diameter"

      By the way can you send me a Test Email? I cannot seem to be able to send emails to your earlier address

      Thank you

      Sumith

      Delete
  4. Sumith PeirisAugust 26, 2015 at 2:58 AM

    Reference my proof for Problem 385, < BED = < EBG = < DGC since BEDG is cyclic and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. MarcoSeptember 10, 2023 at 12:51 AM

    <BHF+<BCF=180
    BHFC is cyclic

    <BEF+<BDF=180
    BEFD is cyclic

    <BEF=<BHF
    BHFE is cyclic

    Combining the above 3 results, BGDFEH are cyclic

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