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
See also:
Complete Problem 464

Level: High School, SAT Prep, College geometry

4 comments:

  1. 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°

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

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  3. 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. 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

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