Monday, May 31, 2010

Problem 464: Square, Center, Arc, Angle, 120 Degrees

Geometry Problem
Click the figure below to see the complete problem 464 about Square, Center, Arc, Angle, 120 Degrees.

Problem 464. Square, Center, Arc, Angle, 120 Degrees
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Complete Problem 464

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 10:08 PM
Labels: , , , , , ,

4 comments:

  1. c .t . e. oJune 1, 2010 at 12:31 AM

    DEC =60° ( DEC equilateral )
    in EGD, DEG = 60°, EDG = 30° ( ADG = 60° - 45°)
    => EGD = 90°
    in EGD, EO, DO bisector ( O center, ODE = 60 - 45° )
    => GO bisector
    => OGD = 45°
    => DOG = 180° - ( 45° + 15°)

    => DOG = 120°

    ReplyDelete
  2. AnonymousJune 2, 2010 at 1:40 AM

    DOC=90deg, DGC=90deg ---> DCG+GOD=180deg.
    DCG=60deg ---> DOG=120deg

    ReplyDelete
  3. Sumith PeirisJanuary 14, 2019 at 5:44 AM

    Tr.s AFD & CDE are equilateral
    So < FDC = 30

    Tr.s BFD & BED are congruent
    So < ODF = 15 = < GCO
    Hence ODCG is concyclic

    So < GOC = < GDC = 30
    Implies < GOD = 120

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sailendra TJanuary 16, 2019 at 12:03 PM

    Connect DE and observe that DEC is an equilateral triangle
    Connect AE and BE and observe that AED congruent with BEC (SAS) => AE=BE
    => BED is congruent to AEC (SSS)
    => m(ECA)=m(BDE)
    => m(GCO)=m(GDO)
    => G,O,D,C are conyclic and since m(GCD)=60 => m(GOD)=120

    ReplyDelete

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