Thursday, August 5, 2010

Problem 492: Cyclic Quadrilateral, Circle, Perpendicular, Parallel

Geometry Problem
Click the figure below to see the complete problem 492 Cyclic Quadrilateral, Circle, Perpendicular, Side, Diagonal, Parallel.

Problem 492. Cyclic Quadrilateral, Circle, Perpendicular, Parallel.
See also:
Complete Problem 492

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 3:46 PM
Labels: , , , , ,

4 comments:

  1. Peter TranAugust 5, 2010 at 10:59 PM

    Denote ( XYZ)= angle XYZ
    1. Connect EC and BG
    Note that E,O,C are collinear ( EAC)=90
    B,O,G are collinear (BDG)=90
    Quadrilateral EGCB have diagonals intersect at midpoints so EGCB is a parallelogram and BC//EG
    2. Note that quadrilateral ADHF is cyclic (FAH)=(FDH)=90
    And (DFH)=(DAH) ( both angles face the same arc DH)
    But (DAH)=(DBC)…………...( both angles face the same arc DC)
    So (DFH)=(DBC) (corresponding angles)
    And BC// FH

    Peter Tran
    vstran@yahoo.com

    ReplyDelete
    Replies
    1. AnonymousOctober 23, 2022 at 12:35 PM

      Hello, Could you please explain this line: (DFH)=(DAH) ( both angles face the same arc DH). I do not see that they have the same arc. Thank you.

      Delete
  2. c .t . e. oAugust 6, 2010 at 12:38 AM

    1) GDC = 90 - BDC, EAB = 90 - BAC => GDC = EAB =>
    arc GC = arc EB => GE // BC
    2) ADHF cyclic =>
    DBC = DAC, DFH = DAC => DBC = DFH as corresponding ang
    => BC // FH

    ReplyDelete
  3. Sumith PeirisJune 22, 2016 at 2:49 AM

    Since ADHF is cyclic < HFD = < HAD = < CBD hence BC//FH

    Also external < H = < EAD = < EGH hence EG //FH

    So FH // BC // EG

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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