Geometry Problem

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## Saturday, January 8, 2011

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

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## Saturday, January 8, 2011

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

Online Geometry theorems, problems, solutions, and related topics.

Click the figure below to see the complete problem 564.

Zoom at: Geometry Problem 564

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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.

ReplyDeleteGNM + GHM =

ReplyDelete180-(90 -A/2 + 90 - D/2) + 180 - (90 - B/2 +90 - C/2)=

A/2 + D/2 + B/2 + C/2 = 180