Tuesday, July 27, 2010

Problem 482: Triangle, Circumcircle, Incenter, Excenter, Midpoint, Cyclic Points

Geometry Problem
Click the figure below to see the complete problem 482 about Triangle, Circumcircle, Incenter, Excenter, Midpoint, Cyclic Points.

Problem 482: Triangle, Circumcircle, Incenter, Excenter, Midpoint, Cyclic Points
See also:
Complete Problem 482

Level: High School, SAT Prep, College geometry

4 comments:

  1. Note that Angle DCE= Angle DBE=90
    Quadrilateral DBEC is cyclic with DE is the diameter of circle DBEC
    Since F is the midpoint of arc BC , F will locate on the perpendicular bisector of chord BC.
    F is the intersecting point of diameter DE and perpen. bisector of chord BC so F must be the center of circle DBEC

    Peter Tran

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  2. Peter: Why DCE = 90

    A,D,F,E lie on the line AE (D & E on bisector,F midpo)
    BF = FC = FE (1)
    ( arc BF = arc FC & BF = FE from P156 )
    ▲DFC isoceles, ang FCD = C/2 + A/2 = 90 - B/2
    ang DFC = ang B (arc AC)
    => ang FDC =180-B-(90-B/2)=90 - B/2
    => ang FDC = ang FCD
    =>FD = FC (2)
    from (1) & (2)
    BF = FC = FE = FD

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  3. CD is angle bisector of (ACB)
    CE is angle bisector of (BCx)
    (ACB)+(BCx)=180
    so (DCE)=90

    Peter Tran

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  4. A,D,F,E are collinear points

    < FDB = < FBD = (A+B)/2 hence FB = FD = FC

    But < AFB = C and < FBE = 90-B/2 - A/2 = C/2 so < FEB also = C/2 hence FE = FB

    SoFD = FB = FE = FC and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete