Saturday, January 8, 2011

Problem 564: Complete Quadrilateral, Incenter, Excenter, Concyclic Points

Geometry Problem
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Problem 564: Complete Quadrilateral, Incenter, Excenter, Concyclic Points.
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Posted by Antonio Gutierrez at 9:23 AM
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2 comments:

  1. EditorJanuary 8, 2011 at 10:28 AM

    Let us trace quadrilateral NMHG, then we note that <AND=(<BAD+<ADC)/2 for external angle bisector reasoning. The same reasoning gives us <BHC=(<ABC+<BCD)/2 and finally adding this angles we get <GNM+<GHM=(sum of internal angles of ADCB)2=360/2=180 and so NMHG is cyclic.

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  2. c .t . e. oJanuary 8, 2011 at 11:29 AM

    GNM + GHM =
    180-(90 -A/2 + 90 - D/2) + 180 - (90 - B/2 +90 - C/2)=
    A/2 + D/2 + B/2 + C/2 = 180

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