Sunday, January 2, 2011

Problem 562: Triangle, Cevian, Incenter, Incircle, Perpendicular, Concyclic points, Cyclic Quadrilateral.

Geometry Problem
Click the figure below to see the complete problem 562.

Problem 562: Triangle, Cevian, Incenter, Incircle, Perpendicular, Concyclic points, Cyclic Quadrilateral.
Zoom at: Geometry Problem 562
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 6:59 AM
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2 comments:

  1. PravinJanuary 2, 2011 at 8:22 AM

    DE, DF bisect the supplementary angles ADB, BDC

    So EDF is a right angle

    By the result of Problem 561, EGF is a right angle

    Hence tha circle on EF as diameter passes through D, G

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  2. Peter TranJanuary 2, 2011 at 8:49 AM

    http://img84.imageshack.us/img84/9576/problem561.png

    Let I is the midpoint of EF and M is the projection of I over AC
    We have angle(EDF)=90 ( see problem 559) and HD=GK (See problem 560)
    M is the midpoint of HK . Since HD=GK so M is also the midpoint of DG and triangle DIG is isosceles.
    And IE=IF=ID=IG .
    Quadrilateral EDGF is cyclic with center of circumcircle at I

    Peter Tran

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