Tuesday, October 13, 2009

Problem 365. Circular Sector of 60 degrees, Midpoints, Perpendicular, Congruence

Proposed Problem
Click the figure below to see the complete problem 365 about Circular Sector of 60 degrees, Midpoints, Perpendicular, Congruence.

 Problem 365. Circular Sector of 60 degrees, Midpoints, Perpendicular, Congruence.
See more:
Complete Geometry Problem 365
Level: High School, SAT Prep, College geometry

7 comments:

  1. GF = 1/2 AB ( as middle line of OAB ) (1)
    GF // AB

    DE = 1/2 AB ( as middle line of ACB ) (2)
    DE // AB

    from (1) & (2)

    GF = De, GF // DE

    at the same way
    GD = EF, GD // EF

    it easy FGDE rombhus

    ReplyDelete
  2. FGDE is a parallelogram per C.t.e.o comment not rhombus.

    ReplyDelete
  3. To Peter
    It is necessary to be rhombus
    OAB equilateral => OC = AB => GD = DE = EF = GF

    ReplyDelete
  4. Midpoints of a quadrilateral connected together always form a parallelogram. But the two diagonals of Quadrilateral ACBO are equal. (Connect OC, revealing OC=OA=OB. Connect AB, revealing Equilateral OAB, so OC=AB.) Quadrilateral ACBO is a rhombus.
    Therefore, DF is perpendicular to EG.

    ReplyDelete
  5. Anonymous
    Thank you for your explanation

    ReplyDelete
  6. OA=OC=OB=AB since Tr. OAB is equilateral

    Then from mid point theorem GDEF is a rhombus whose diagonals are perpendicular to each other

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete