Saturday, August 3, 2013

Problem 906: Bicentric Quadrilateral, Incircle, Circumcircle, Circunscribed, Inscribed, Perpendicular

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 906.

Online Geometry Problem 906: Bicentric Quadrilateral, Incircle, Circumcircle, Circunscribed, Inscribed, Perpendicular.

Posted by Antonio Gutierrez at 12:32 PM
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2 comments:

  1. UnknownAugust 3, 2013 at 2:55 PM

    A=a
    C=180-a
    FHA=BFH=b [=FGH]
    AEG=EGD=c [=EFG]
    FH meet EG at K
    EKH=360-a-b-c=FKG
    CGK=180-EGD=180-c
    CFK=180-b
    360=180-a+180-b+180-c+360-a-b-c
    a+b+c=270
    FKE=90

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  2. Jacob HA (EMK2000)August 3, 2013 at 5:58 PM

    Let EG and FH meet at K.

    Then
    ∠FKG
    = 1/2 ( ∠FIG + ∠EIH )
    = 1/2 ((180 - ∠FCG ) + (180 - ∠EAH))
    = 90 °

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