Saturday, June 4, 2016

Geometry Problem 1223: Circle, Parallel Chords

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

Click the figure below to view more details of problem 1223.

Geometry Problem 1223: Circle, Parallel Chords

Posted by Antonio Gutierrez at 7:53 PM
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4 comments:

  1. APOSTOLIS MANOLOUDISJune 4, 2016 at 9:53 PM

    Is arc AC=arcBD=arcEF. ThereforeCE//AF.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS PIRAEUS GREECE

    ReplyDelete
  2. Sumith PeirisJune 4, 2016 at 10:56 PM

    BDFE is a concyclic trapezoid, hence

    < EBD = < FEB = < ECD = @ say ....(1)

    Similarly ABDC is a concyclic trapezoid, hence < ACD = < CDB = CEB = @ say ...(2)

    From (1) and (2) < CEF = < ACE

    But < ACE and < AFE are supplementary

    Hence < CEF and < AFE are supplementary implying that AF//CE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Sumith PeirisJune 4, 2016 at 10:59 PM

    Another method........

    Since BDFE and ABDC are concyclic trapezoids AC = BD = EF

    So in concyclic ACEF, AC = EF and the result follows

    ReplyDelete
  4. c.t.e.oJune 4, 2016 at 11:31 PM

    arc AC = arc BD, arc EF = arc BD
    => arc EF = arc AC

    ReplyDelete

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